🎯 Section A: Objective & MCQs (1 Mark Each)
1 Mark
CBSE 2022 Term-1
Q1. A negatively charged body $X$ experiences electrostatic repulsion from another charged body $Y$. An uncharged or charged body $Z$ is brought near $Y$ and experiences electrostatic attraction towards $Y$. Which of the following statements represents the correct state of body $Z$?
1 Mark
CBSE 2022 Term-1
Q2. In a Millikan-type chamber experiment, three microscopic latex spheres carry initial electric charges of $+3e$, $+5e$, and $-3e$ respectively. The three spheres are momentarily brought into simultaneous physical contact and then separated. Which of the following charge combinations is physically possible on the separated spheres?
1 Mark
CBSE 2022 Term-1
Q3. An object initially possesses a net positive charge of $+1.0\text{ C}$. It then gains $5.0 \times 10^{18}$ electrons. The final net charge on the object becomes:
1 Mark
CBSE 2022 Term-1
Q4. Three charges $q, -q$, and $q_0$ are placed on coordinate axes as shown below: $+q$ is at $(a, 0)$, $-q$ is at $(0, a)$, and test charge $q_0$ is fixed at origin $(0, 0)$. Taking $k = \frac{1}{4\pi\varepsilon_0}$, determine the net electrostatic force on $q_0$.
1 Mark
CBSE CFPQ Competency Question
Q5. Two insulated dielectric rods with opposite static charges ($+q$ and $-q$) at their extremities are pivoted at their centers. They can rotate freely in the plane. In which of the given orientations are the rods in a state of stable equilibrium (such that they return to original orientation when slightly displaced)?
⚖️ Section B: Assertion & Reason (1 Mark Each)
Directions: (a) Both A and R are true and R is correct explanation of A. (b) Both A and R are true but R is NOT correct explanation. (c) A is true but R is false. (d) A is false but R is true.
1 Mark
CBSE Board Standard
Assertion (A): The magnitude of electric charge on an object is strictly relativistic invariant.
Reason (R): The measured electric charge does not depend on the relative velocity of the particle or the frame of reference.
Reason (R): The measured electric charge does not depend on the relative velocity of the particle or the frame of reference.
1 Mark
CBSE Board Standard
Assertion (A): Electrostatic attraction can exist even between a charged body and an uncharged (neutral) body.
Reason (R): In Coulomb attraction, both interacting bodies must always carry opposite and equal charges.
Reason (R): In Coulomb attraction, both interacting bodies must always carry opposite and equal charges.
📝 Section C: Short Answer Questions (2 Marks Each)
2 Marks
CBSE Core Syllabus
Q1. Define and explain two fundamental intrinsic properties of electric charges.
1. Quantisation of Charge (1 Mark): Electric charge exists in discrete packets rather than a continuous spectrum. Any observable charge $q$ is always an integral multiple of basic quantum $e$:
$$q = \pm ne \quad (\text{where } n = 1, 2, 3\dots \text{ and } e = 1.6 \times 10^{-19}\text{ C})$$
2. Conservation of Charge (1 Mark): The net electric charge in an isolated physical system remains invariant over time. Charge can neither be created nor destroyed; it can only transfer from one object to another.
2 Marks
NCERT Intext / CBSE Board
Q2. If $10^9$ electrons move out of a neutral body to another body every second, calculate the total time required to accumulate a net negative charge of $1.0\text{ C}$ on the receiving body.
Step 1: Charge transferred per second (1 Mark):
$$q_{\text{per sec}} = n \cdot e = 10^9 \times (1.6 \times 10^{-19}\text{ C}) = 1.6 \times 10^{-10}\text{ C/s}$$
Step 2: Total time calculation (1 Mark):
$$t = \frac{Q}{q_{\text{per sec}}} = \frac{1\text{ C}}{1.6 \times 10^{-10}\text{ C/s}} = 6.25 \times 10^9\text{ seconds}$$
Converting into years:
$$t = \frac{6.25 \times 10^9}{365 \times 24 \times 3600} \approx \mathbf{198.18\text{ Years}}$$
2 Marks
NCERT Exemplar / CBSE
Q3. Estimate the total amount of positive (and negative) charge present in a standard cup ($250\text{ g}$) of pure water.
Step 1: Number of water molecules in 250 g (1 Mark):
Molecular mass of $H_2O = 18\text{ g/mol}$. $$N = \left(\frac{250}{18}\right) \times N_A = \left(\frac{250}{18}\right) \times 6.023 \times 10^{23} \approx 8.365 \times 10^{24}\text{ molecules}$$
Molecular mass of $H_2O = 18\text{ g/mol}$. $$N = \left(\frac{250}{18}\right) \times N_A = \left(\frac{250}{18}\right) \times 6.023 \times 10^{23} \approx 8.365 \times 10^{24}\text{ molecules}$$
Step 2: Net Charge calculation (1 Mark):
One $H_2O$ molecule has 2 protons (from H) + 8 protons (from O) = 10 protons (and 10 electrons). $$Q = N \times 10 \times e = (8.365 \times 10^{24}) \times 10 \times (1.6 \times 10^{-19}\text{ C}) = \mathbf{1.34 \times 10^7\text{ C}}$$
One $H_2O$ molecule has 2 protons (from H) + 8 protons (from O) = 10 protons (and 10 electrons). $$Q = N \times 10 \times e = (8.365 \times 10^{24}) \times 10 \times (1.6 \times 10^{-19}\text{ C}) = \mathbf{1.34 \times 10^7\text{ C}}$$
3 Marks
NCERT Intext / CBSE Core
Q3. A small conducting charged sphere $A$ is suspended by an insulating nylon thread. Another identical conducting sphere $B$ held by an insulating handle is brought close to $A$ such that the separation between their centers is $r_1 = 10\text{ cm}$. The repulsive electrostatic force $F$ is recorded.
Identical uncharged spheres $C$ and $D$ are then brought in contact with $A$ and $B$ respectively and then removed. Sphere $B$ is now moved closer to $A$ such that the separation becomes $r_2 = 5.0\text{ cm}$. Calculate the new electrostatic force of repulsion $F'$ between $A$ and $B$ in terms of $F$.
Identical uncharged spheres $C$ and $D$ are then brought in contact with $A$ and $B$ respectively and then removed. Sphere $B$ is now moved closer to $A$ such that the separation becomes $r_2 = 5.0\text{ cm}$. Calculate the new electrostatic force of repulsion $F'$ between $A$ and $B$ in terms of $F$.
Step 1: Initial Force Expression (1 Mark):
Let sphere $A$ have initial charge $q$ and sphere $B$ have charge $q'$. At separation $r = 10\text{ cm}$: $$F = \frac{1}{4\pi\varepsilon_0}\frac{q q'}{r^2}$$
Let sphere $A$ have initial charge $q$ and sphere $B$ have charge $q'$. At separation $r = 10\text{ cm}$: $$F = \frac{1}{4\pi\varepsilon_0}\frac{q q'}{r^2}$$
Step 2: Charge redistribution after contact (1 Mark):
When identical uncharged sphere $C$ touches $A$, the charge divides equally: $$q_A' = \frac{q + 0}{2} = \frac{q}{2}$$ Similarly, when identical uncharged sphere $D$ touches $B$: $$q_B' = \frac{q' + 0}{2} = \frac{q'}{2}$$
When identical uncharged sphere $C$ touches $A$, the charge divides equally: $$q_A' = \frac{q + 0}{2} = \frac{q}{2}$$ Similarly, when identical uncharged sphere $D$ touches $B$: $$q_B' = \frac{q' + 0}{2} = \frac{q'}{2}$$
Step 3: New Force Calculation at distance $r' = \frac{r}{2} = 5\text{ cm}$ (1 Mark):
$$F' = \frac{1}{4\pi\varepsilon_0}\frac{q_A' q_B'}{(r')^2} = \frac{1}{4\pi\varepsilon_0}\frac{\left(\frac{q}{2}\right)\left(\frac{q'}{2}\right)}{\left(\frac{r}{2}\right)^2} = \frac{1}{4\pi\varepsilon_0}\frac{\frac{q q'}{4}}{\frac{r^2}{4}} = \frac{1}{4\pi\varepsilon_0}\frac{q q'}{r^2} = \mathbf{F}$$ Conclusion: The electrostatic force of repulsion remains completely unchanged (Unite/Unaltered = F).
$$F' = \frac{1}{4\pi\varepsilon_0}\frac{q_A' q_B'}{(r')^2} = \frac{1}{4\pi\varepsilon_0}\frac{\left(\frac{q}{2}\right)\left(\frac{q'}{2}\right)}{\left(\frac{r}{2}\right)^2} = \frac{1}{4\pi\varepsilon_0}\frac{\frac{q q'}{4}}{\frac{r^2}{4}} = \frac{1}{4\pi\varepsilon_0}\frac{q q'}{r^2} = \mathbf{F}$$ Conclusion: The electrostatic force of repulsion remains completely unchanged (Unite/Unaltered = F).
3 Marks
NCERT Intext / CBSE Board
Q4. Three charges $+q$, $+q$, and $-q$ are placed at vertices $A$, $B$, and $C$ of an equilateral triangle of side $l$. Determine the magnitude and vector direction of the resultant electrostatic force acting on each charge. Also verify whether the sum of all forces on the three charges equals zero.
Step 1: Magnitude of Pairwise Force (1 Mark):
The electrostatic interaction between any pair of charges has the same magnitude: $$F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{l^2}$$
The electrostatic interaction between any pair of charges has the same magnitude: $$F = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{l^2}$$
Step 2: Net Force on Vertex A ($+q$) and Vertex B ($+q$) (1 Mark):
• At $A$: Repelled by $+q$ at $B$ ($\vec{F}_{AB}$ along $BA$) and attracted by $-q$ at $C$ ($\vec{F}_{AC}$ along $AC$). Angle between them is $180^\circ - 60^\circ = 120^\circ$. $$F_1 = \sqrt{F^2 + F^2 + 2F^2\cos 120^\circ} = \sqrt{2F^2 - F^2} = \mathbf{F} = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{l^2}$$ (Directed along the direction bisecting the angle, parallel to side $BC$).
• By exact symmetry, net force at vertex $B$ is also: $\mathbf{F_2 = F}$ (parallel to side $AC$).
• At $A$: Repelled by $+q$ at $B$ ($\vec{F}_{AB}$ along $BA$) and attracted by $-q$ at $C$ ($\vec{F}_{AC}$ along $AC$). Angle between them is $180^\circ - 60^\circ = 120^\circ$. $$F_1 = \sqrt{F^2 + F^2 + 2F^2\cos 120^\circ} = \sqrt{2F^2 - F^2} = \mathbf{F} = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{l^2}$$ (Directed along the direction bisecting the angle, parallel to side $BC$).
• By exact symmetry, net force at vertex $B$ is also: $\mathbf{F_2 = F}$ (parallel to side $AC$).
Step 3: Net Force on Vertex C ($-q$) (1 Mark):
• At $C$: Attracted by $+q$ at $A$ along $CA$ and attracted by $+q$ at $B$ along $CB$. Angle between these forces is $60^\circ$. $$F_3 = \sqrt{F^2 + F^2 + 2F^2\cos 60^\circ} = \sqrt{2F^2 + F^2} = \mathbf{\sqrt{3}F} = \mathbf{\frac{\sqrt{3}}{4\pi\varepsilon_0}\frac{q^2}{l^2}}$$ (Directed along the angle bisector vertically downwards).
Verification of Newton's Third Law: $$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \mathbf{0}$$
• At $C$: Attracted by $+q$ at $A$ along $CA$ and attracted by $+q$ at $B$ along $CB$. Angle between these forces is $60^\circ$. $$F_3 = \sqrt{F^2 + F^2 + 2F^2\cos 60^\circ} = \sqrt{2F^2 + F^2} = \mathbf{\sqrt{3}F} = \mathbf{\frac{\sqrt{3}}{4\pi\varepsilon_0}\frac{q^2}{l^2}}$$ (Directed along the angle bisector vertically downwards).
Verification of Newton's Third Law: $$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \mathbf{0}$$
5 Marks
CBSE Board Exam Classic
Q2. Two identical positive point charges $+q$ each are held fixed at a distance of $2\text{ m}$ apart in air. A third point charge $Q$ of unknown sign and magnitude is placed along the line joining them such that the entire three-charge system remains in complete electrostatic equilibrium.
(i) Determine the exact position of charge $Q$ from either charge.
(ii) Prove whether the charge $Q$ must be positive or negative.
(iii) Calculate the exact magnitude of charge $Q$ in terms of $q$.
(i) Determine the exact position of charge $Q$ from either charge.
(ii) Prove whether the charge $Q$ must be positive or negative.
(iii) Calculate the exact magnitude of charge $Q$ in terms of $q$.
Part (i): Position of Charge Q (2 Marks):
Let charge $Q$ be placed at distance $x$ from charge $A$ ($+q$). Distance from charge $B$ ($+q$) is $(2 - x)$.
For $Q$ to be in equilibrium, the electrostatic forces on $Q$ due to $A$ and $B$ must be equal in magnitude and opposite in direction: $$\frac{1}{4\pi\varepsilon_0}\frac{q Q}{x^2} = \frac{1}{4\pi\varepsilon_0}\frac{q Q}{(2 - x)^2} \implies x^2 = (2 - x)^2$$ Taking square root on both sides: $$x = 2 - x \implies 2x = 2 \implies \mathbf{x = 1\text{ m}}$$ Hence, charge $Q$ must be placed exactly at the midpoint ($1\text{ m}$) between the two $+q$ charges.
Let charge $Q$ be placed at distance $x$ from charge $A$ ($+q$). Distance from charge $B$ ($+q$) is $(2 - x)$.
For $Q$ to be in equilibrium, the electrostatic forces on $Q$ due to $A$ and $B$ must be equal in magnitude and opposite in direction: $$\frac{1}{4\pi\varepsilon_0}\frac{q Q}{x^2} = \frac{1}{4\pi\varepsilon_0}\frac{q Q}{(2 - x)^2} \implies x^2 = (2 - x)^2$$ Taking square root on both sides: $$x = 2 - x \implies 2x = 2 \implies \mathbf{x = 1\text{ m}}$$ Hence, charge $Q$ must be placed exactly at the midpoint ($1\text{ m}$) between the two $+q$ charges.
Part (ii): Nature and Sign of Charge Q (1 Mark):
If $Q$ were positive, both $+q$ and $Q$ would push the charge at $A$ in the same direction towards the left, making equilibrium impossible.
Therefore, charge $Q$ must be negative so that its attractive force balances the mutual repulsion between the two $+q$ charges.
If $Q$ were positive, both $+q$ and $Q$ would push the charge at $A$ in the same direction towards the left, making equilibrium impossible.
Therefore, charge $Q$ must be negative so that its attractive force balances the mutual repulsion between the two $+q$ charges.
Part (iii): Magnitude of Charge Q (2 Marks):
For the entire system to be in equilibrium, net force on charge at $A$ must also be zero: $$F_{\text{net on } A} = F_{AB} + F_{AO} = 0$$ $$\frac{1}{4\pi\varepsilon_0}\frac{q \cdot q}{(2)^2} + \frac{1}{4\pi\varepsilon_0}\frac{q \cdot Q}{(1)^2} = 0$$ Canceling common factor $\frac{q}{4\pi\varepsilon_0}$: $$\frac{q}{4} + Q = 0 \implies \mathbf{Q = -\frac{q}{4}}$$ Final Conclusion: The third charge must have a magnitude of $\mathbf{\frac{q}{4}}$ and a negative sign, placed at the midpoint $\mathbf{1\text{ m}}$ from either charge.
For the entire system to be in equilibrium, net force on charge at $A$ must also be zero: $$F_{\text{net on } A} = F_{AB} + F_{AO} = 0$$ $$\frac{1}{4\pi\varepsilon_0}\frac{q \cdot q}{(2)^2} + \frac{1}{4\pi\varepsilon_0}\frac{q \cdot Q}{(1)^2} = 0$$ Canceling common factor $\frac{q}{4\pi\varepsilon_0}$: $$\frac{q}{4} + Q = 0 \implies \mathbf{Q = -\frac{q}{4}}$$ Final Conclusion: The third charge must have a magnitude of $\mathbf{\frac{q}{4}}$ and a negative sign, placed at the midpoint $\mathbf{1\text{ m}}$ from either charge.
🏆 Section E: 5-Mark Long Exam Numericals
5 Marks
CBSE Board Standard
Q1. Two identical point charges $+q$ each are separated by a distance of $2\text{ m}$ in air. A third point charge $Q$ of unknown magnitude and sign is placed on the line joining them such that the entire three-charge system remains in electrostatic equilibrium.
(i) Find the position of charge $Q$.
(ii) Determine the magnitude and sign of charge $Q$.
(i) Find the position of charge $Q$.
(ii) Determine the magnitude and sign of charge $Q$.
Part (i): Position of Third Charge $Q$ (2 Marks):
Let $Q$ be placed at distance $x$ from one charge. Distance from second charge is $(2 - x)$. For $Q$ to be in equilibrium: $$\frac{1}{4\pi\varepsilon_0}\frac{qQ}{x^2} = \frac{1}{4\pi\varepsilon_0}\frac{qQ}{(2 - x)^2} \implies x^2 = (2 - x)^2$$ $$x = 2 - x \implies 2x = 2 \implies \mathbf{x = 1\text{ m}}$$ Hence, charge $Q$ must be placed exactly at the midpoint ($1\text{ m}$) between the two charges.
Let $Q$ be placed at distance $x$ from one charge. Distance from second charge is $(2 - x)$. For $Q$ to be in equilibrium: $$\frac{1}{4\pi\varepsilon_0}\frac{qQ}{x^2} = \frac{1}{4\pi\varepsilon_0}\frac{qQ}{(2 - x)^2} \implies x^2 = (2 - x)^2$$ $$x = 2 - x \implies 2x = 2 \implies \mathbf{x = 1\text{ m}}$$ Hence, charge $Q$ must be placed exactly at the midpoint ($1\text{ m}$) between the two charges.
Part (ii): Sign and Magnitude of Charge $Q$ (3 Marks):
For the entire system to be in equilibrium, net force on one of the outer $+q$ charges must also be zero: $$F_{\text{net on outer } q} = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{(2)^2} + \frac{1}{4\pi\varepsilon_0}\frac{qQ}{(1)^2} = 0$$ $$\frac{q^2}{4} + qQ = 0 \implies qQ = -\frac{q^2}{4} \implies \mathbf{Q = -\frac{q}{4}}$$ Conclusion: The charge $Q$ must be negative in nature with magnitude $\mathbf{q/4}$.
For the entire system to be in equilibrium, net force on one of the outer $+q$ charges must also be zero: $$F_{\text{net on outer } q} = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{(2)^2} + \frac{1}{4\pi\varepsilon_0}\frac{qQ}{(1)^2} = 0$$ $$\frac{q^2}{4} + qQ = 0 \implies qQ = -\frac{q^2}{4} \implies \mathbf{Q = -\frac{q}{4}}$$ Conclusion: The charge $Q$ must be negative in nature with magnitude $\mathbf{q/4}$.
🎯 Section A: Electric Field & Dipole Very Short answer (1 Mark Each)
1 MarkCBSE 2022 (Re)
1. The magnitude of electric field due to a point charge $2q$, at distance $r$ is $E$. Then the magnitude of electric field due to a uniformly charged thin spherical shell of radius $R$ with total charge $q$ at a distance $\frac{r}{2}\ (r \gg R)$ will be:
1 MarkCBSE 2022 (Ev)
2. A square sheet of side '$a$' is lying parallel to $XY$ plane at $z = a$. The electric field in the region is $\vec{E} = cz^2\hat{k}$. The electric flux through the sheet is:
1 MarkCBSE SQP 2023 (Ev)
3. An electric dipole placed in an electric field of intensity $2 \times 10^5\text{ N/C}$ at an angle of $30^\circ$ experiences a torque equal to $4\text{ Nm}$. The charge on the dipole of dipole length $2\text{ cm}$ is:
1 MarkCBSE CFPQ 2023 (Ev)
4. Electric field lines are pictorial representations of electric fields due to static charges on the plane of a paper. Study the given electric field representation and identify one INCORRECT qualitative impression given by this representation:
1 MarkCBSE Board (Ev)
5. If the intensity of an electric field is given by $\vec{E} = 8\hat{i} + 8\hat{j} + \hat{k}$ in $\text{NC}^{-1}$. The electric flux through a surface $\vec{S} = 10\hat{i}\text{ m}^2$ is:
1 MarkCBSE (Ev)
6. An electric dipole consists of two charges of magnitude placed $0.1\,\mu\text{C}$ separated by a distance of $2\text{ cm}$. The dipole is in an external field of $10^5\text{ N/C}$. What maximum torque does the electric field exert on the dipole?
⚖️ Section B: Assertion and Reason (1 Mark Each)
1 MarkCBSE 2022 (Re)
1. Assertion (A): A negative charge in an electric field starts to move along the direction of the electric field.
Reason (R): On a negative charge, a force acts in the opposite direction of the electric field.
Reason (R): On a negative charge, a force acts in the opposite direction of the electric field.
1 MarkCBSE CFPQ 2022 (Re)
2. Assertion (A): An electric dipole is in stable equilibrium when placed in a uniform electric field with its dipole moment opposite to the field.
Reason (R): No torque acts on an electric dipole when its dipole moment is in a direction opposite to the field.
Reason (R): No torque acts on an electric dipole when its dipole moment is in a direction opposite to the field.
🎯 Section A: Electric Field & Dipole Short answer (2 Mark Each)
2 MarksCBSE 2020 (Un)
1. The work done in moving a charge particle between two points in a uniform electric field, does not depend on the path followed by the particle. Why?
Because electrostatic field is a conservative field (1 Mark). Work done depends exclusively on the initial and final positions: $W = q\Delta V$ (1 Mark).
2 MarksCBSE 2019 (Cr)
2. Draw the pattern of electric field lines, when a point charge $-Q$ is kept near an uncharged conducting plate.
Positive and negative charges get induced on the metal surface due to electrostatic induction (1 Mark). Field lines terminate perpendicularly onto the conducting plate surface (1 Mark).
2 MarksCBSE Board Exam
Q3. (b) If the net electric flux through a closed surface is zero, what can be said about the net charge enclosed inside that Gaussian surface?
By Gauss's Theorem, total flux $\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}$ (1 Mark).
If net flux is zero, it implies that the **net algebraic charge enclosed inside the Gaussian surface is strictly zero** ($q_{\text{enclosed}} = 0$), though equal and opposite charges may exist inside (1 Mark).
If net flux is zero, it implies that the **net algebraic charge enclosed inside the Gaussian surface is strictly zero** ($q_{\text{enclosed}} = 0$), though equal and opposite charges may exist inside (1 Mark).
2 MarksCBSE Board Conceptual
Q3. What does the statement, "The electric field of a point charge exhibits spherical symmetry, while that of an electric dipole displays cylindrical symmetry," signify?
1. Point Charge (Spherical Symmetry - 1 Mark): The field magnitude $E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$ remains constant at all points on the surface of a sphere of radius $r$ centered around the charge. It is independent of direction, giving spherical symmetry.
2. Electric Dipole (Cylindrical Symmetry - 1 Mark): The field remains uniform at all locations situated along a cylindrical surface of radius $r$ aligned symmetrically along the dipole's axis. Its appearance is consistent within all planes intersecting the dipole axis, hence possessing cylindrical symmetry.
📝 Section D: Short Answer Type Questions (3 Marks Each)
3 MarksCBSE Board Exam (Ev)
Q1. Two charges of value $2\,\mu\text{C}$ and $-50\,\mu\text{C}$ are placed $80\text{ cm}$ apart. Calculate the distance of the point from the smaller charge where the net electric field intensity is zero.
Step 1: Conceptual Position Analysis (1 Mark):
Since charges have opposite signs, the neutral point where $E = 0$ lies outside on the line joining them, on the side of the smaller charge ($2\,\mu\text{C}$). Let distance be $x$.
Since charges have opposite signs, the neutral point where $E = 0$ lies outside on the line joining them, on the side of the smaller charge ($2\,\mu\text{C}$). Let distance be $x$.
Step 2: Mathematical Equating (1 Mark):
$$\frac{k q_1}{x^2} = \frac{k |q_2|}{(80 + x)^2} \implies \frac{2 \times 10^{-6}}{x^2} = \frac{50 \times 10^{-6}}{(80 + x)^2}$$ $$\frac{1}{x^2} = \frac{25}{(80 + x)^2} \implies \frac{1}{x} = \frac{5}{80 + x}$$
$$\frac{k q_1}{x^2} = \frac{k |q_2|}{(80 + x)^2} \implies \frac{2 \times 10^{-6}}{x^2} = \frac{50 \times 10^{-6}}{(80 + x)^2}$$ $$\frac{1}{x^2} = \frac{25}{(80 + x)^2} \implies \frac{1}{x} = \frac{5}{80 + x}$$
Step 3: Final Calculation (1 Mark):
$$80 + x = 5x \implies 4x = 80 \implies \mathbf{x = 20\text{ cm}}$$ The intensity is zero at a distance of **$20\text{ cm}$** from the smaller charge.
$$80 + x = 5x \implies 4x = 80 \implies \mathbf{x = 20\text{ cm}}$$ The intensity is zero at a distance of **$20\text{ cm}$** from the smaller charge.
3 MarksCBSE Board Exam
Q2. Calculate the maximum torque experienced by a water molecule whose electric dipole moment is $6.2 \times 10^{-30}\text{ C}\cdot\text{m}$, when it is placed in a uniform electric field of intensity $10^6\text{ N/C}$.
Given Data (1 Mark):
Dipole moment $p = 6.2 \times 10^{-30}\text{ C}\cdot\text{m}$, Electric field $E = 10^6\text{ N/C}$.
Dipole moment $p = 6.2 \times 10^{-30}\text{ C}\cdot\text{m}$, Electric field $E = 10^6\text{ N/C}$.
Formula & Calculation (2 Marks):
Torque formula: $\tau = p E \sin\theta$. For maximum torque, $\theta = 90^\circ \implies \sin 90^\circ = 1$. $$\tau_{\text{max}} = (6.2 \times 10^{-30}\text{ C}\cdot\text{m}) \times (10^6\text{ N/C}) = \mathbf{6.2 \times 10^{-24}\text{ N}\cdot\text{m}}$$
Torque formula: $\tau = p E \sin\theta$. For maximum torque, $\theta = 90^\circ \implies \sin 90^\circ = 1$. $$\tau_{\text{max}} = (6.2 \times 10^{-30}\text{ C}\cdot\text{m}) \times (10^6\text{ N/C}) = \mathbf{6.2 \times 10^{-24}\text{ N}\cdot\text{m}}$$
3 MarksNCERT Intext / CBSE Board
Q3. Two point charges $q_1$ and $q_2$ of magnitudes $+10^{-8}\text{ C}$ and $-10^{-8}\text{ C}$ respectively, are placed $0.1\text{ m}$ apart. Calculate the electric fields at points $A$ (midpoint), $B$ (outside left), and $C$ (equator apex) as shown in the figure.
1. Electric Field at Midpoint A ($r = 0.05\text{ m}$ - 1 Mark):
$E_{1A}$ due to $q_1$ and $E_{2A}$ due to $q_2$ both point towards the right.
$$E_A = E_{1A} + E_{2A} = 2 \times \left(\frac{9 \times 10^9 \times 10^{-8}}{(0.05)^2}\right) = \mathbf{7.2 \times 10^4\text{ N/C}} \quad (\text{towards right})$$
$E_{1A}$ due to $q_1$ and $E_{2A}$ due to $q_2$ both point towards the right.
$$E_A = E_{1A} + E_{2A} = 2 \times \left(\frac{9 \times 10^9 \times 10^{-8}}{(0.05)^2}\right) = \mathbf{7.2 \times 10^4\text{ N/C}} \quad (\text{towards right})$$
2. Electric Field at Point B ($r = 0.05\text{ m}$ to left of $q_1$ - 1 Mark):
$$E_{1B} = \frac{9 \times 10^9 \times 10^{-8}}{(0.05)^2} = 3.6 \times 10^4\text{ N/C} \quad (\leftarrow)$$ $$E_{2B} = \frac{9 \times 10^9 \times 10^{-8}}{(0.15)^2} = 4 \times 10^3\text{ N/C} \quad (\rightarrow)$$ $$E_B = E_{1B} - E_{2B} = \mathbf{3.2 \times 10^4\text{ N/C}} \quad (\text{towards left})$$
$$E_{1B} = \frac{9 \times 10^9 \times 10^{-8}}{(0.05)^2} = 3.6 \times 10^4\text{ N/C} \quad (\leftarrow)$$ $$E_{2B} = \frac{9 \times 10^9 \times 10^{-8}}{(0.15)^2} = 4 \times 10^3\text{ N/C} \quad (\rightarrow)$$ $$E_B = E_{1B} - E_{2B} = \mathbf{3.2 \times 10^4\text{ N/C}} \quad (\text{towards left})$$
3. Electric Field at Apex Point C (1 Mark):
Equilateral triangle geometry yields individual vector magnitudes of $9 \times 10^3\text{ N/C}$ inclined at $60^\circ$. By vector addition: $\mathbf{E_C = 9 \times 10^3\text{ N/C}}$ (directed parallel to $q_1 q_2$ towards the right).
Equilateral triangle geometry yields individual vector magnitudes of $9 \times 10^3\text{ N/C}$ inclined at $60^\circ$. By vector addition: $\mathbf{E_C = 9 \times 10^3\text{ N/C}}$ (directed parallel to $q_1 q_2$ towards the right).
🏆 Section D: Long Answer Type Questions (5 Marks Each)
5 Marks
CBSE Board Exam (2019)
Q1. (a) Derive an analytical expression for the electric field intensity $\vec{E}$ at any point on the equatorial line (perpendicular bisector) of an electric dipole of dipole moment $\vec{p}$.
(b) Two point charges $q_A = 3\,\mu\text{C}$ and $q_B = -3\,\mu\text{C}$ are located $20\text{ cm}$ apart in vacuum. Calculate the total electric field at the midpoint $O$ of the line $AB$ joining the two charges.
(b) Two point charges $q_A = 3\,\mu\text{C}$ and $q_B = -3\,\mu\text{C}$ are located $20\text{ cm}$ apart in vacuum. Calculate the total electric field at the midpoint $O$ of the line $AB$ joining the two charges.
Part (a): Equatorial Line Derivation (3 Marks):
• Let charges $-q$ and $+q$ be separated by $2a$. Point $P$ lies on the perpendicular bisector at distance $r$.
• Distance from each charge to $P$ is $R = \sqrt{r^2 + a^2}$. The magnitudes of electric fields are: $$E_{+q} = E_{-q} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2 + a^2}$$ • Resolving components: Vertical components $E\sin\theta$ are equal and opposite, hence they cancel out. Horizontal components $E\cos\theta$ reinforce each other along a direction anti-parallel to $\vec{p}$: $$E_{\text{eq}} = 2E\cos\theta = 2 \left(\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2+a^2}\right) \left(\frac{a}{\sqrt{r^2+a^2}}\right) = \mathbf{\frac{1}{4\pi\varepsilon_0}\frac{p}{(r^2+a^2)^{3/2}}}$$ For a short dipole ($r \gg a$), $a^2$ is neglected: $\vec{E}_{\text{eq}} = -\frac{1}{4\pi\varepsilon_0}\frac{\vec{p}}{r^3}$.
• Let charges $-q$ and $+q$ be separated by $2a$. Point $P$ lies on the perpendicular bisector at distance $r$.
• Distance from each charge to $P$ is $R = \sqrt{r^2 + a^2}$. The magnitudes of electric fields are: $$E_{+q} = E_{-q} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2 + a^2}$$ • Resolving components: Vertical components $E\sin\theta$ are equal and opposite, hence they cancel out. Horizontal components $E\cos\theta$ reinforce each other along a direction anti-parallel to $\vec{p}$: $$E_{\text{eq}} = 2E\cos\theta = 2 \left(\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2+a^2}\right) \left(\frac{a}{\sqrt{r^2+a^2}}\right) = \mathbf{\frac{1}{4\pi\varepsilon_0}\frac{p}{(r^2+a^2)^{3/2}}}$$ For a short dipole ($r \gg a$), $a^2$ is neglected: $\vec{E}_{\text{eq}} = -\frac{1}{4\pi\varepsilon_0}\frac{\vec{p}}{r^3}$.
Part (b): Numerical at Midpoint $O$ (2 Marks):
• Given: $q_A = 3\,\mu\text{C} = 3 \times 10^{-6}\text{ C}$, $q_B = -3\,\mu\text{C} = -3 \times 10^{-6}\text{ C}$, total distance $AB = 20\text{ cm} = 0.2\text{ m}$.
• Distance from midpoint $O$ to each charge is $r = 10\text{ cm} = 0.1\text{ m}$.
• Electric field at $O$ due to $q_A$ (positive charge points away from $A$, i.e., towards $B$): $$E_A = \frac{9 \times 10^9 \times 3 \times 10^{-6}}{(0.1)^2} = \frac{27 \times 10^3}{0.01} = 2.7 \times 10^6\text{ N/C} \quad (\text{towards } B)$$ • Electric field at $O$ due to $q_B$ (negative charge points towards $B$): $$E_B = \frac{9 \times 10^9 \times 3 \times 10^{-6}}{(0.1)^2} = 2.7 \times 10^6\text{ N/C} \quad (\text{towards } B)$$ • Since both vectors act in the same direction (towards $B$): $$E_{\text{resultant}} = E_A + E_B = 2.7 \times 10^6 + 2.7 \times 10^6 = \mathbf{5.4 \times 10^6\text{ N/C}} \quad (\text{from } A \text{ to } B)$$
• Given: $q_A = 3\,\mu\text{C} = 3 \times 10^{-6}\text{ C}$, $q_B = -3\,\mu\text{C} = -3 \times 10^{-6}\text{ C}$, total distance $AB = 20\text{ cm} = 0.2\text{ m}$.
• Distance from midpoint $O$ to each charge is $r = 10\text{ cm} = 0.1\text{ m}$.
• Electric field at $O$ due to $q_A$ (positive charge points away from $A$, i.e., towards $B$): $$E_A = \frac{9 \times 10^9 \times 3 \times 10^{-6}}{(0.1)^2} = \frac{27 \times 10^3}{0.01} = 2.7 \times 10^6\text{ N/C} \quad (\text{towards } B)$$ • Electric field at $O$ due to $q_B$ (negative charge points towards $B$): $$E_B = \frac{9 \times 10^9 \times 3 \times 10^{-6}}{(0.1)^2} = 2.7 \times 10^6\text{ N/C} \quad (\text{towards } B)$$ • Since both vectors act in the same direction (towards $B$): $$E_{\text{resultant}} = E_A + E_B = 2.7 \times 10^6 + 2.7 \times 10^6 = \mathbf{5.4 \times 10^6\text{ N/C}} \quad (\text{from } A \text{ to } B)$$
5 Marks
CBSE Board Exam Classic
Q2. (a) An electric dipole consisting of charges $\pm q$ separated by $2a$ is placed. Deduce an expression for electric field $\vec{E}$ at distance $x$ on its axial line. Hence show that for $x \gg a$, $\vec{E} \to \frac{2\vec{p}}{4\pi\varepsilon_0 x^3}$.
(b) Given electric field in a region is $\vec{E} = 2x\hat{i}$, find the net electric flux through a cube of side $a$ placed in this region and determine the total charge enclosed by it. OR.
(c) Define torque acting on a dipole in uniform field and explain behavior in non-uniform field.
(b) Given electric field in a region is $\vec{E} = 2x\hat{i}$, find the net electric flux through a cube of side $a$ placed in this region and determine the total charge enclosed by it. OR.
(c) Define torque acting on a dipole in uniform field and explain behavior in non-uniform field.
Part (a): Axial Line Derivation (3 Marks):
• Let observation point be at distance $x$ from the center on the axis.
• Net axial field is sum of fields due to $+q$ and $-q$: $$E_{\text{axial}} = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{(x-a)^2} - \frac{1}{(x+a)^2} \right] = \frac{1}{4\pi\varepsilon_0}\frac{q \cdot (4ax)}{(x^2 - a^2)^2} = \frac{1}{4\pi\varepsilon_0}\frac{2(q \cdot 2a)x}{(x^2 - a^2)^2} = \frac{1}{4\pi\varepsilon_0}\frac{2px}{(x^2 - a^2)^2}$$ • For short dipole ($x \gg a$): $(x^2 - a^2)^2 \approx x^4$, hence: $$\vec{E}_{\text{axial}} = \mathbf{\frac{1}{4\pi\varepsilon_0}\frac{2\vec{p}}{x^3}}$$
• Let observation point be at distance $x$ from the center on the axis.
• Net axial field is sum of fields due to $+q$ and $-q$: $$E_{\text{axial}} = \frac{q}{4\pi\varepsilon_0} \left[ \frac{1}{(x-a)^2} - \frac{1}{(x+a)^2} \right] = \frac{1}{4\pi\varepsilon_0}\frac{q \cdot (4ax)}{(x^2 - a^2)^2} = \frac{1}{4\pi\varepsilon_0}\frac{2(q \cdot 2a)x}{(x^2 - a^2)^2} = \frac{1}{4\pi\varepsilon_0}\frac{2px}{(x^2 - a^2)^2}$$ • For short dipole ($x \gg a$): $(x^2 - a^2)^2 \approx x^4$, hence: $$\vec{E}_{\text{axial}} = \mathbf{\frac{1}{4\pi\varepsilon_0}\frac{2\vec{p}}{x^3}}$$
Part (b): Cube Flux & Enclosed Charge (2 Marks):
• Field is $\vec{E} = 2x\hat{i}$. Only the faces perpendicular to the $x$-axis contribute to the flux.
• At left face ($x = 0$): $\Phi_{\text{left}} = 0$ since $E = 0$.
• At right face ($x = a$): $\vec{E} = 2a\hat{i}$ and area vector $\vec{A} = a^2\hat{i}$. Thus $\Phi_{\text{right}} = (2a)(a^2) = 2a^3$.
• Total net flux $\Phi = \Phi_{\text{right}} - \Phi_{\text{left}} = \mathbf{2a^3}$.
• By Gauss's Law, enclosed charge $q_{\text{enc}} = \varepsilon_0 \Phi = \mathbf{2a^3\varepsilon_0}$.
• Field is $\vec{E} = 2x\hat{i}$. Only the faces perpendicular to the $x$-axis contribute to the flux.
• At left face ($x = 0$): $\Phi_{\text{left}} = 0$ since $E = 0$.
• At right face ($x = a$): $\vec{E} = 2a\hat{i}$ and area vector $\vec{A} = a^2\hat{i}$. Thus $\Phi_{\text{right}} = (2a)(a^2) = 2a^3$.
• Total net flux $\Phi = \Phi_{\text{right}} - \Phi_{\text{left}} = \mathbf{2a^3}$.
• By Gauss's Law, enclosed charge $q_{\text{enc}} = \varepsilon_0 \Phi = \mathbf{2a^3\varepsilon_0}$.
📐 Continuous Charge Distribution & Gauss's Law (PYQs)
1 Mark
CBSE 2020
Q1. If the net electric flux passing through a closed Gaussian surface is strictly zero, then we can infer that:
1 Mark
CBSE ASQP 2023
Q2. An infinite line of charge has a linear charge density of $\lambda = 10^{-7}\text{ C/m}$. What will be the magnitude of the electrostatic force acting on an alpha particle placed at a perpendicular distance of $4\text{ cm}$ ($0.04\text{ m}$) from the line of charge?
1 Mark
CBSE CFPQ 2023
Q3. For a closed Gaussian surface through which the net electric flux is zero, analyze the following statements:
P. No charges reside inside the Gaussian surface.
Q. The net charge inside the surface is zero.
R. The electric field is zero everywhere on the surface.
S. The number of field lines entering the surface is equal to the number of lines exiting the surface.
Identify which of the statements COULD BE TRUE:
P. No charges reside inside the Gaussian surface.
Q. The net charge inside the surface is zero.
R. The electric field is zero everywhere on the surface.
S. The number of field lines entering the surface is equal to the number of lines exiting the surface.
Identify which of the statements COULD BE TRUE:
1 Mark
CBSE CFPQ 2023
Q4. A lightning rod is constructed from conducting material with sharp metal spikes on top connected to an earthing wire. Which statement from below DOES NOT contribute to the correct working principle of a lightning conductor?
📐 Assertion-Reasoning & Gauss's Law Applications (PYQs)
1 Mark Each
CBSE Board Standard
1. Assertion (A): With the help of Gauss's theorem, we can find electric field at any point.
Reason (R): Gauss's theorem cannot be applied for any type of charge distribution.
Reason (R): Gauss's theorem cannot be applied for any type of charge distribution.
1 Mark
CBSE Board Standard
2. Assertion (A): The electric flux emanating out and entering a closed surface are $8 \times 10^3\text{ V}\cdot\text{m}$ and $2 \times 10^3\text{ V}\cdot\text{m}$ respectively. The net charge enclosed by the surface is $0.053\,\mu\text{C}$.
Reason (R): Gauss's theorem in electrostatics may be applied to verify this.
Reason (R): Gauss's theorem in electrostatics may be applied to verify this.
📝 Subjective Questions (Very Short & Short Answer)
2 Marks
CBSE Board Exam
Q1. A point charge causes an electric flux of $-3 \times 10^{-16}\text{ N}\cdot\text{m}^2/\text{C}$ to pass through a spherical Gaussian surface.
(a) Calculate the value of the point charge.
(b) If the radius of the Gaussian surface is doubled, how much flux would pass through the surface?
(a) Calculate the value of the point charge.
(b) If the radius of the Gaussian surface is doubled, how much flux would pass through the surface?
(a) Value of Point Charge (1 Mark):
By Gauss's theorem, $\Phi = \frac{q}{\varepsilon_0} \implies q = \Phi \cdot \varepsilon_0$. $$q = (-3 \times 10^{-16}) \times (8.854 \times 10^{-12}) = \mathbf{-2.655 \times 10^{-27}\text{ C}}$$
By Gauss's theorem, $\Phi = \frac{q}{\varepsilon_0} \implies q = \Phi \cdot \varepsilon_0$. $$q = (-3 \times 10^{-16}) \times (8.854 \times 10^{-12}) = \mathbf{-2.655 \times 10^{-27}\text{ C}}$$
(b) Effect of doubling radius (1 Mark):
The electric flux passing through a closed surface depends solely on the total net charge enclosed by the surface, not on its size or radius. Therefore, the flux remains **unchanged ($-\mathbf{3 \times 10^{-16}\text{ N}\cdot\text{m}^2/\text{C}}$)**.
The electric flux passing through a closed surface depends solely on the total net charge enclosed by the surface, not on its size or radius. Therefore, the flux remains **unchanged ($-\mathbf{3 \times 10^{-16}\text{ N}\cdot\text{m}^2/\text{C}}$)**.
2 Marks
CBSE Board Exam
Q2. A metallic spherical shell has an inner radius $R_1$ and outer radius $R_2$. A charge $Q$ is placed at the center of the spherical cavity. What will be the surface charge density on (i) the inner surface, and (ii) the outer surface?
(i) Inner Surface Charge Density (1 Mark):
Due to induction, a charge of $-Q$ is induced on the inner surface. Inner surface area is $4\pi R_1^2$. $$\sigma_{\text{inner}} = \mathbf{-\frac{Q}{4\pi R_1^2}}$$
Due to induction, a charge of $-Q$ is induced on the inner surface. Inner surface area is $4\pi R_1^2$. $$\sigma_{\text{inner}} = \mathbf{-\frac{Q}{4\pi R_1^2}}$$
(ii) Outer Surface Charge Density (1 Mark):
An equal and opposite charge $+Q$ is induced on the outer surface. Outer surface area is $4\pi R_2^2$. $$\sigma_{\text{outer}} = \mathbf{+\frac{Q}{4\pi R_2^2}}$$
An equal and opposite charge $+Q$ is induced on the outer surface. Outer surface area is $4\pi R_2^2$. $$\sigma_{\text{outer}} = \mathbf{+\frac{Q}{4\pi R_2^2}}$$
3 Marks
CBSE Board Exam (Important)
Q3. A small metal sphere carrying charge $+Q$ is located at the center of a spherical cavity inside a large uncharged metallic spherical shell as shown. Use Gauss's law to find expressions for the electric field at points $P_1$ (inside cavity at distance $r_1$) and $P_2$ (outside shell at distance $r_2$).
1. Electric Field at Point $P_1$ (Distance $r_1$ inside cavity - 1.5 Marks):
Applying Gaussian sphere of radius $r_1$ around center charge $+Q$: $$\oint \vec{E}_1 \cdot d\vec{A} = E_1 (4\pi r_1^2) = \frac{Q}{\varepsilon_0} \implies \mathbf{E_1 = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r_1^2}}$$
Applying Gaussian sphere of radius $r_1$ around center charge $+Q$: $$\oint \vec{E}_1 \cdot d\vec{A} = E_1 (4\pi r_1^2) = \frac{Q}{\varepsilon_0} \implies \mathbf{E_1 = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r_1^2}}$$
2. Electric Field at Point $P_2$ (Distance $r_2$ outside shell - 1.5 Marks):
The total net charge enclosed by a Gaussian sphere of radius $r_2$ passing through point $P_2$ is $+Q$ (since induced charges on inner/outer shell surfaces cancel out algebraically). $$\oint \vec{E}_2 \cdot d\vec{A} = E_2 (4\pi r_2^2) = \frac{Q}{\varepsilon_0} \implies \mathbf{E_2 = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r_2^2}}$$
The total net charge enclosed by a Gaussian sphere of radius $r_2$ passing through point $P_2$ is $+Q$ (since induced charges on inner/outer shell surfaces cancel out algebraically). $$\oint \vec{E}_2 \cdot d\vec{A} = E_2 (4\pi r_2^2) = \frac{Q}{\varepsilon_0} \implies \mathbf{E_2 = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r_2^2}}$$
📐 Gauss's Law Applications & Charged Sheets (PYQs)
3 Marks
CBSE Board Exam (2020)
Q1. (a) A hollow conducting sphere of inner radius $r_1$ and outer radius $r_2$ has a charge $Q$ on its surface. A point charge $q$ is also placed at the center of the sphere.
(i) What is the surface charge density on the inner and outer surface of the sphere?
(ii) Use Gauss' law of electrostatics to obtain the expression for the electric field at a point lying outside the sphere.
OR
(b) (i) An infinitely long thin straight wire has uniform linear charge density $\lambda$. Obtain the expression for electric field $E$ at distance $x$ from the wire, using Gauss' law.
(ii) Show graphically the variation of this electric field $E$ as a function of distance $x$ from the wire.
(i) What is the surface charge density on the inner and outer surface of the sphere?
(ii) Use Gauss' law of electrostatics to obtain the expression for the electric field at a point lying outside the sphere.
OR
(b) (i) An infinitely long thin straight wire has uniform linear charge density $\lambda$. Obtain the expression for electric field $E$ at distance $x$ from the wire, using Gauss' law.
(ii) Show graphically the variation of this electric field $E$ as a function of distance $x$ from the wire.
(a) Hollow Conducting Sphere:
• (i) Due to charge $q$ at center, charge induced on inner surface is $-q$ and on outer surface is $Q + q$.
Surface charge density: $\sigma_{\text{inner}} = \frac{-q}{4\pi r_1^2}$, $\sigma_{\text{outer}} = \frac{Q + q}{4\pi r_2^2}$.
• (ii) For a point outside at distance $r$ ($r > r_2$), total net enclosed charge is $Q + q$. By Gauss Law: $$E (4\pi r^2) = \frac{Q + q}{\varepsilon_0} \implies \mathbf{E = \frac{1}{4\pi\varepsilon_0}\frac{Q + q}{r^2}}$$
• (i) Due to charge $q$ at center, charge induced on inner surface is $-q$ and on outer surface is $Q + q$.
Surface charge density: $\sigma_{\text{inner}} = \frac{-q}{4\pi r_1^2}$, $\sigma_{\text{outer}} = \frac{Q + q}{4\pi r_2^2}$.
• (ii) For a point outside at distance $r$ ($r > r_2$), total net enclosed charge is $Q + q$. By Gauss Law: $$E (4\pi r^2) = \frac{Q + q}{\varepsilon_0} \implies \mathbf{E = \frac{1}{4\pi\varepsilon_0}\frac{Q + q}{r^2}}$$
(b) Infinitely Long Straight Wire (OR part):
• (i) Considering a cylindrical Gaussian surface of radius $x$ and length $l$: $$\oint \vec{E} \cdot d\vec{A} = E \cdot (2\pi x l) = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{\lambda l}{\varepsilon_0} \implies \mathbf{E = \frac{\lambda}{2\pi\varepsilon_0 x}}$$ • (ii) Graph of $E$ versus $x$ shows a rectangular hyperbola curve where $E \propto \frac{1}{x}$.
• (i) Considering a cylindrical Gaussian surface of radius $x$ and length $l$: $$\oint \vec{E} \cdot d\vec{A} = E \cdot (2\pi x l) = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{\lambda l}{\varepsilon_0} \implies \mathbf{E = \frac{\lambda}{2\pi\varepsilon_0 x}}$$ • (ii) Graph of $E$ versus $x$ shows a rectangular hyperbola curve where $E \propto \frac{1}{x}$.
3 Marks
CBSE Board Exam (2019)
Q2. Two large charged plane sheets of charge densities $+\sigma$ and $-2\sigma\text{ C/m}^2$ are arranged vertically with a separation $d$ between them. Deduce expressions for the electric field at points:
(i) to the left of the first sheet,
(ii) to the right of the second sheet, and
(iii) between the two sheets.
OR
(b) A spherical conducting shell of inner radius $r_1$ and outer radius $r_2$ has a charge $Q$.
(i) Find surface charge density on inner and outer surfaces.
(ii) Is electric field inside a cavity (with no charge) zero, independent of shell shape? Explain.
(i) to the left of the first sheet,
(ii) to the right of the second sheet, and
(iii) between the two sheets.
OR
(b) A spherical conducting shell of inner radius $r_1$ and outer radius $r_2$ has a charge $Q$.
(i) Find surface charge density on inner and outer surfaces.
(ii) Is electric field inside a cavity (with no charge) zero, independent of shell shape? Explain.
Two Charged Sheets Regions:
Electric field due to single sheet of charge density $\sigma$ is $E = \frac{\sigma}{2\varepsilon_0}$.
• (i) Region I (Left of Sheet 1): Fields point outwards from $+\sigma$ and inwards to $-2\sigma$: $$E_I = -\frac{\sigma}{2\varepsilon_0} - \frac{2\sigma}{2\varepsilon_0} = \mathbf{-\frac{3\sigma}{2\varepsilon_0}} \quad (\text{towards left})$$ • (ii) Region II (Right of Sheet 2): $$E_{II} = +\frac{\sigma}{2\varepsilon_0} + \frac{2\sigma}{2\varepsilon_0} = \mathbf{+\frac{3\sigma}{2\varepsilon_0}} \quad (\text{towards right})$$ • (iii) Region III (Between sheets): $$E_{III} = \frac{\sigma}{2\varepsilon_0} - \frac{2\sigma}{2\varepsilon_0} = \mathbf{-\frac{\sigma}{2\varepsilon_0}}$$
Electric field due to single sheet of charge density $\sigma$ is $E = \frac{\sigma}{2\varepsilon_0}$.
• (i) Region I (Left of Sheet 1): Fields point outwards from $+\sigma$ and inwards to $-2\sigma$: $$E_I = -\frac{\sigma}{2\varepsilon_0} - \frac{2\sigma}{2\varepsilon_0} = \mathbf{-\frac{3\sigma}{2\varepsilon_0}} \quad (\text{towards left})$$ • (ii) Region II (Right of Sheet 2): $$E_{II} = +\frac{\sigma}{2\varepsilon_0} + \frac{2\sigma}{2\varepsilon_0} = \mathbf{+\frac{3\sigma}{2\varepsilon_0}} \quad (\text{towards right})$$ • (iii) Region III (Between sheets): $$E_{III} = \frac{\sigma}{2\varepsilon_0} - \frac{2\sigma}{2\varepsilon_0} = \mathbf{-\frac{\sigma}{2\varepsilon_0}}$$
OR Part (Spherical Shell):
• (i) Inner surface charge density = $0$ (since no cavity charge). Outer surface charge density = $\frac{Q}{4\pi r_2^2}$.
• (ii) Yes, electric field inside an empty cavity of a conductor is always zero due to electrostatic shielding, completely independent of the outer shape of the shell.
• (i) Inner surface charge density = $0$ (since no cavity charge). Outer surface charge density = $\frac{Q}{4\pi r_2^2}$.
• (ii) Yes, electric field inside an empty cavity of a conductor is always zero due to electrostatic shielding, completely independent of the outer shape of the shell.
📐 Gauss's Law Applications & Circular Orbit Derivations (PYQs)
5 Marks
CBSE Board Exam (2023)
Q1. (a) Use Gauss' law to obtain an expression for the electric field due to an infinitely long thin straight wire with uniform linear charge density $\lambda$.
(b) An infinitely long positively charged straight wire has a linear charge density $\lambda$. An electron is revolving in a circle with a constant speed $v$ such that the wire passes through the center, and is perpendicular to the plane of the circle. Find the kinetic energy of the electron in terms of magnitude of its charge ($e$) and linear charge density $\lambda$ on the wire.
(c) Draw a graph of kinetic energy as a function of linear charge density $\lambda$.
(b) An infinitely long positively charged straight wire has a linear charge density $\lambda$. An electron is revolving in a circle with a constant speed $v$ such that the wire passes through the center, and is perpendicular to the plane of the circle. Find the kinetic energy of the electron in terms of magnitude of its charge ($e$) and linear charge density $\lambda$ on the wire.
(c) Draw a graph of kinetic energy as a function of linear charge density $\lambda$.
(a) Electric Field Derivation (2 Marks):
• Consider a cylindrical Gaussian surface of radius $r$ and length $l$ around the infinite wire.
• Flux through flat end faces is zero (since $\vec{E} \perp d\vec{A}$). Flux passes only through curved surface: $$\oint \vec{E} \cdot d\vec{A} = E \cdot (2\pi r l) = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{\lambda l}{\varepsilon_0}$$ $$E \cdot (2\pi r l) = \frac{\lambda l}{\varepsilon_0} \implies \mathbf{E = \frac{\lambda}{2\pi\varepsilon_0 r}}$$
• Consider a cylindrical Gaussian surface of radius $r$ and length $l$ around the infinite wire.
• Flux through flat end faces is zero (since $\vec{E} \perp d\vec{A}$). Flux passes only through curved surface: $$\oint \vec{E} \cdot d\vec{A} = E \cdot (2\pi r l) = \frac{q_{\text{enclosed}}}{\varepsilon_0} = \frac{\lambda l}{\varepsilon_0}$$ $$E \cdot (2\pi r l) = \frac{\lambda l}{\varepsilon_0} \implies \mathbf{E = \frac{\lambda}{2\pi\varepsilon_0 r}}$$
(b) Kinetic Energy of Revolving Electron (2 Marks):
• The electrostatic force provides the necessary centripetal force for circular motion: $$F_e = \frac{m v^2}{r} \implies e \cdot E = \frac{m v^2}{r}$$ • Substituting $E = \frac{\lambda}{2\pi\varepsilon_0 r}$: $$e \left(\frac{\lambda}{2\pi\varepsilon_0 r}\right) = \frac{m v^2}{r} \implies m v^2 = \frac{e \lambda}{2\pi\varepsilon_0}$$ • Kinetic Energy ($K$) is given by $K = \frac{1}{2}m v^2$: $$K = \frac{1}{2} \left(\frac{e \lambda}{2\pi\varepsilon_0}\right) = \mathbf{\frac{e \lambda}{4\pi\varepsilon_0}}$$ Notice that the radius $r$ cancels out completely, showing that kinetic energy is independent of the orbital radius.
• The electrostatic force provides the necessary centripetal force for circular motion: $$F_e = \frac{m v^2}{r} \implies e \cdot E = \frac{m v^2}{r}$$ • Substituting $E = \frac{\lambda}{2\pi\varepsilon_0 r}$: $$e \left(\frac{\lambda}{2\pi\varepsilon_0 r}\right) = \frac{m v^2}{r} \implies m v^2 = \frac{e \lambda}{2\pi\varepsilon_0}$$ • Kinetic Energy ($K$) is given by $K = \frac{1}{2}m v^2$: $$K = \frac{1}{2} \left(\frac{e \lambda}{2\pi\varepsilon_0}\right) = \mathbf{\frac{e \lambda}{4\pi\varepsilon_0}}$$ Notice that the radius $r$ cancels out completely, showing that kinetic energy is independent of the orbital radius.
(c) Graph of Kinetic Energy vs $\lambda$ (1 Mark):
Since $K = \left(\frac{e}{4\pi\varepsilon_0}\right) \lambda$, the graph of Kinetic Energy ($K$) versus linear charge density ($\lambda$) is a straight line passing through the origin with a positive slope.
Since $K = \left(\frac{e}{4\pi\varepsilon_0}\right) \lambda$, the graph of Kinetic Energy ($K$) versus linear charge density ($\lambda$) is a straight line passing through the origin with a positive slope.
📐 Spherical Shell Derivations & Flux Numericals (PYQs)
5 Marks
CBSE Board Exam (2020)
Q1. (a) Using Gauss' law, derive the expression for electric field due to a spherical shell of uniform charge distribution $\sigma$ and radius $R$ at a point lying at a distance $x$ from the center of the shell, such that:
(i) $0 < x < R$, and (ii) $x > R$.
(b) An electric field is uniform and acts along $+x$ direction for $x > 0$ and along $-x$ direction for $x < 0$ with magnitude $E = 200\text{ N/C}$. A right circular cylinder of length $20\text{ cm}$ ($0.2\text{ m}$) and radius $5\text{ cm}$ ($0.05\text{ m}$) is centered at the origin along $x$-axis (flat faces at $x = +10\text{ cm}$ and $x = -10\text{ cm}$). Find:
(i) The net outward flux through the cylinder.
(ii) The net charge present inside the cylinder.
(i) $0 < x < R$, and (ii) $x > R$.
(b) An electric field is uniform and acts along $+x$ direction for $x > 0$ and along $-x$ direction for $x < 0$ with magnitude $E = 200\text{ N/C}$. A right circular cylinder of length $20\text{ cm}$ ($0.2\text{ m}$) and radius $5\text{ cm}$ ($0.05\text{ m}$) is centered at the origin along $x$-axis (flat faces at $x = +10\text{ cm}$ and $x = -10\text{ cm}$). Find:
(i) The net outward flux through the cylinder.
(ii) The net charge present inside the cylinder.
(a) Spherical Shell Derivations (3 Marks):
• (i) Inside the shell ($0 < x < R$): A Gaussian sphere of radius $r < R$ encloses zero net charge ($q_{\text{in}} = 0$). By Gauss's Law, $\oint \vec{E} \cdot d\vec{A} = \frac{0}{\varepsilon_0} \implies \mathbf{E = 0}$.
• (ii) Outside the shell ($x > R$): Total charge on shell $q = \sigma \cdot (4\pi R^2)$. Applying Gaussian sphere of radius $r = x$: $$E \cdot (4\pi x^2) = \frac{q}{\varepsilon_0} \implies \mathbf{E = \frac{1}{4\pi\varepsilon_0}\frac{q}{x^2} = \frac{\sigma R^2}{\varepsilon_0 x^2}}$$
• (i) Inside the shell ($0 < x < R$): A Gaussian sphere of radius $r < R$ encloses zero net charge ($q_{\text{in}} = 0$). By Gauss's Law, $\oint \vec{E} \cdot d\vec{A} = \frac{0}{\varepsilon_0} \implies \mathbf{E = 0}$.
• (ii) Outside the shell ($x > R$): Total charge on shell $q = \sigma \cdot (4\pi R^2)$. Applying Gaussian sphere of radius $r = x$: $$E \cdot (4\pi x^2) = \frac{q}{\varepsilon_0} \implies \mathbf{E = \frac{1}{4\pi\varepsilon_0}\frac{q}{x^2} = \frac{\sigma R^2}{\varepsilon_0 x^2}}$$
(b) Cylinder Numerical Calculation (2 Marks):
• (i) Net outward flux ($\Phi$): Field is directed outwards on both flat circular faces of area $A = \pi r^2 = 3.14 \times (0.05)^2 = 0.00785\text{ m}^2$.
$$\Phi = 2 \cdot (E \cdot A) = 2 \times (200\text{ N/C}) \times (3.14 \times 0.0025\text{ m}^2) = \mathbf{3.14\text{ N}\cdot\text{m}^2/\text{C}}$$ • (ii) Net Charge Enclosed ($Q_{\text{enc}}$): $$Q_{\text{enc}} = \varepsilon_0 \cdot \Phi = (8.854 \times 10^{-12}) \times 3.14 = \mathbf{2.78 \times 10^{-11}\text{ C}}$$
• (i) Net outward flux ($\Phi$): Field is directed outwards on both flat circular faces of area $A = \pi r^2 = 3.14 \times (0.05)^2 = 0.00785\text{ m}^2$.
$$\Phi = 2 \cdot (E \cdot A) = 2 \times (200\text{ N/C}) \times (3.14 \times 0.0025\text{ m}^2) = \mathbf{3.14\text{ N}\cdot\text{m}^2/\text{C}}$$ • (ii) Net Charge Enclosed ($Q_{\text{enc}}$): $$Q_{\text{enc}} = \varepsilon_0 \cdot \Phi = (8.854 \times 10^{-12}) \times 3.14 = \mathbf{2.78 \times 10^{-11}\text{ C}}$$
1 Mark
CBSE Board Exam
Q2. Draw a clean, precise graph showing the variation of Kinetic Energy ($K$) of the revolving electron as a function of the linear charge density ($\lambda$) on the straight wire.
Explanation:
From the derivation $K = \frac{e\lambda}{4\pi\varepsilon_0}$, kinetic energy $K$ is directly proportional to linear charge density $\lambda$ ($K \propto \lambda$). Hence, the graph is a straight line passing through the origin with a constant positive slope $\frac{e}{4\pi\varepsilon_0}$.
From the derivation $K = \frac{e\lambda}{4\pi\varepsilon_0}$, kinetic energy $K$ is directly proportional to linear charge density $\lambda$ ($K \propto \lambda$). Hence, the graph is a straight line passing through the origin with a constant positive slope $\frac{e}{4\pi\varepsilon_0}$.
📐 Electric Field Outside a Spherical Shell & Field Lines (PYQs)
5 Marks
CBSE Board Exam (Standard)
Q1. (a) Apply Gauss's law to derive an expression for the electric field strength at any location outside a uniformly charged thin spherical shell with a radius of $R$ and a charge density of $\sigma\text{ C/m}^2$.
(b) Illustrate the field lines when the sphere's charge density is (i) positive and (ii) negative.
(b) Illustrate the field lines when the sphere's charge density is (i) positive and (ii) negative.
(a) Derivation for Outside Point ($r > R$) (3 Marks):
• Let a thin spherical shell of radius $R$ carry total charge $q$. To find electric field at an outside point $P$ at distance $r$ from center $O$, we construct a concentric spherical Gaussian surface of radius $r$.
• At every point on the Gaussian surface, electric field $\vec{E}$ is directed radially outward, parallel to area vector $d\vec{A}$ ($\theta = 0^\circ$).
• According to Gauss's Theorem: $$\oint \vec{E} \cdot d\vec{A} = \frac{q}{\varepsilon_0} \implies E \cdot (4\pi r^2) = \frac{q}{\varepsilon_0}$$ $$E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$$ • In terms of surface charge density $\sigma$ ($q = \sigma \cdot 4\pi R^2$): $$\mathbf{E = \frac{\sigma R^2}{\varepsilon_0 r^2} \quad \text{or} \quad \vec{E} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{n}}$$
• Let a thin spherical shell of radius $R$ carry total charge $q$. To find electric field at an outside point $P$ at distance $r$ from center $O$, we construct a concentric spherical Gaussian surface of radius $r$.
• At every point on the Gaussian surface, electric field $\vec{E}$ is directed radially outward, parallel to area vector $d\vec{A}$ ($\theta = 0^\circ$).
• According to Gauss's Theorem: $$\oint \vec{E} \cdot d\vec{A} = \frac{q}{\varepsilon_0} \implies E \cdot (4\pi r^2) = \frac{q}{\varepsilon_0}$$ $$E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}$$ • In terms of surface charge density $\sigma$ ($q = \sigma \cdot 4\pi R^2$): $$\mathbf{E = \frac{\sigma R^2}{\varepsilon_0 r^2} \quad \text{or} \quad \vec{E} = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{n}}$$
(b) Field Lines Illustration (2 Marks):
• (i) Positive Charge Density ($\sigma > 0$): Electric field lines radiate outwards uniformly in all directions from the spherical surface.
• (ii) Negative Charge Density ($\sigma < 0$): Electric field lines point radially inwards perpendicularly toward the spherical surface from infinity.
• (i) Positive Charge Density ($\sigma > 0$): Electric field lines radiate outwards uniformly in all directions from the spherical surface.
• (ii) Negative Charge Density ($\sigma < 0$): Electric field lines point radially inwards perpendicularly toward the spherical surface from infinity.
3 Marks
CBSE Board Exam Numerical
Q2. Determine the following for a uniformly charged conducting sphere with a diameter of $2.5\text{ m}$ and a surface charge density of $100\,\mu\text{C/m}^2$:
(i) The total charge present on the sphere.
(ii) The overall electric flux passing through the sphere.
(i) The total charge present on the sphere.
(ii) The overall electric flux passing through the sphere.
Given Data:
• Diameter $D = 2.5\text{ m} \implies \text{Radius } R = \frac{D}{2} = \frac{2.5}{2} = 1.25\text{ m}$.
• Surface charge density $\sigma = 100\,\mu\text{C/m}^2 = 100 \times 10^{-6}\text{ C/m}^2 = 10^{-4}\text{ C/m}^2$.
• Diameter $D = 2.5\text{ m} \implies \text{Radius } R = \frac{D}{2} = \frac{2.5}{2} = 1.25\text{ m}$.
• Surface charge density $\sigma = 100\,\mu\text{C/m}^2 = 100 \times 10^{-6}\text{ C/m}^2 = 10^{-4}\text{ C/m}^2$.
(i) Total Charge Present on the Sphere ($q$ - 1.5 Marks):
Formula: $q = \sigma \cdot A = \sigma \cdot (4\pi R^2)$ $$q = (10^{-4}\text{ C/m}^2) \times 4 \times 3.14 \times (1.25\text{ m})^2$$ $$q = 10^{-4} \times 12.56 \times 1.5625 = \mathbf{1.96 \times 10^{-3}\text{ C} \quad (\text{or } 1.96\text{ mC})}$$
Formula: $q = \sigma \cdot A = \sigma \cdot (4\pi R^2)$ $$q = (10^{-4}\text{ C/m}^2) \times 4 \times 3.14 \times (1.25\text{ m})^2$$ $$q = 10^{-4} \times 12.56 \times 1.5625 = \mathbf{1.96 \times 10^{-3}\text{ C} \quad (\text{or } 1.96\text{ mC})}$$
(ii) Overall Electric Flux Passing Through the Sphere ($\Phi$ - 1.5 Marks):
By Gauss's Law, total flux is determined by the total enclosed charge: $$\Phi = \frac{q}{\varepsilon_0} = \frac{1.96 \times 10^{-3}\text{ C}}{8.854 \times 10^{-12}\text{ C}^2/\text{N}\cdot\text{m}^2}$$ $$\Phi \approx \mathbf{2.2 \times 10^8\text{ N}\cdot\text{m}^2/\text{C}}$$
By Gauss's Law, total flux is determined by the total enclosed charge: $$\Phi = \frac{q}{\varepsilon_0} = \frac{1.96 \times 10^{-3}\text{ C}}{8.854 \times 10^{-12}\text{ C}^2/\text{N}\cdot\text{m}^2}$$ $$\Phi \approx \mathbf{2.2 \times 10^8\text{ N}\cdot\text{m}^2/\text{C}}$$
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Disclaimer & Cross-Check
Special Advice & Warning for Students:
These notes and step-by-step PYQ solutions have been prepared with utmost care and accuracy, yet human error during typing or mathematical calculation is possible. Therefore, before accepting any numerical or formula as absolute truth, be sure to cross-check with your NCERT textbook and teachers. Shorveer Notes is not liable for any unexpected calculation errors. Use your own discretion and judgment.
These notes and step-by-step PYQ solutions have been prepared with utmost care and accuracy, yet human error during typing or mathematical calculation is possible. Therefore, before accepting any numerical or formula as absolute truth, be sure to cross-check with your NCERT textbook and teachers. Shorveer Notes is not liable for any unexpected calculation errors. Use your own discretion and judgment.