Appeared in: 2025 (3 sets)
(a) (i) A small conducting sphere A of radius $r$ charged to a potential $V$, is enclosed by a spherical conducting shell B of radius $R$. If A and B are connected by a thin wire, calculate the final potential on sphere A and shell B.
(ii) Write two characteristics of equipotential surfaces. A uniform electric field of $50 \text{ NC}^{-1}$ is set up in a region along $+x$ axis. If the potential at the origin $(0, 0)$ is $220 \text{ V}$, find the potential at a point $(4\text{m}, 3\text{m})$.
OR
(b) (i) What is difference between an open surface and a closed surface ? Draw elementary surface vector $d\vec{S}$ for a spherical surface $S$.
(ii) Define electric flux through a surface. Give the significance of a Gaussian surface. A charge outside a Gaussian surface does not contribute to total electric flux through the surface. Why ?
(iii) A small spherical shell $S_1$ has point charges $q_1 = -3 \text{ }\mu\text{C}$, $q_2 = -2 \text{ }\mu\text{C}$ and $q_3 = 9 \text{ }\mu\text{C}$ inside it. This shell is enclosed by another big spherical shell $S_2$. A point charge $Q$ is placed in between the two surfaces $S_1$ and $S_2$. If the electric flux through the surface $S_2$ is four times the flux through surface $S_1$, find charge $Q$.
✓ Verified CBSE Marking Scheme
Official Evaluation Steps & EL10 Insight
Solution (a)
(i) Potential of connected spheres: [1.5 Marks]
- Step 1 (Governing Principle): When two conductors are connected by a wire, charge flows from higher potential to lower potential until both reach a common potential. Since the inner sphere A is inside the shell B, all charge $q$ on A will flow to the outer surface of B.
- Step 2 (Mathematical Working): Initial charge on A is $q = C_A V = (4\pi\epsilon_0 r)V$. After connection, the entire charge $q$ resides on the outer surface of shell B. The potential of the system (both A and B) is determined by the potential at the surface of B: $$V_{final} = \frac{1}{4\pi\epsilon_0} \frac{q}{R}$$
- Step 3 (Final Answer): Substituting $q$: $$V_{final} = \frac{1}{4\pi\epsilon_0} \frac{(4\pi\epsilon_0 r)V}{R} = V\left(\frac{r}{R}\right)$$
(ii) Equipotential Surfaces and Potential Calculation: [1.5 Marks]
- Characteristics: (1) No work is done in moving a test charge between two points on an equipotential surface. (2) Electric field lines are always normal to the equipotential surface.
- Step 1 (Formula): The relation between electric field and potential is $dV = -\vec{E} \cdot d\vec{r}$. For a uniform field: $V_B - V_A = -\vec{E} \cdot (\vec{r}_B - \vec{r}_A)$.
- Step 2 (Working): Given $\vec{E} = 50\hat{i} \text{ NC}^{-1}$, $V_{(0,0)} = 220 \text{ V}$, and displacement $\Delta\vec{r} = (4-0)\hat{i} + (3-0)\hat{j} = 4\hat{i} + 3\hat{j}$. $$V_{(4,3)} - V_{(0,0)} = -(50\hat{i}) \cdot (4\hat{i} + 3\hat{j}) = -200 \text{ V}$$
- Step 3 (Final Answer): $V_{(4,3)} = 220 - 200 = 20 \text{ V}$.
💡 EL10 Concept Insight: Think of the electric field like a downhill slope; if you walk sideways along the hill (equipotential), you don't go up or down, but if you walk in the direction of the field, you lose potential energy just like walking down a hill.
Solution (b)
(i) Surfaces and Vector Representation: [1 Mark]
- Difference: An open surface (like a flat disk) has a boundary, whereas a closed surface (like a sphere) encloses a volume and has no boundary.
- Drawing: For a spherical surface, the elementary area vector $d\vec{S}$ is directed radially outward, perpendicular to the surface at every point.
(ii) Electric Flux and Gaussian Surface: [1 Mark]
- Definition: Electric flux is the total number of electric field lines passing normally through a given surface area: $\phi_E = \int \vec{E} \cdot d\vec{S}$.
- Significance: A Gaussian surface is a hypothetical closed