CBSE 2025, Set 2

Question

(a) (i) The electric field in a region is given by $\vec{E} = 40x \hat{i}	ext{ N/C}$. Find the amount of work done in taking a unit positive charge from a point $(0, 3	ext{m})$ to the point $(5	ext{m}, 0)$.
(ii) A charge $Q$ is distributed over two concentric hollow spheres of radii $r$ and $R (R > r)$ such that their surface charge densities are equal. Find: (I) the electric field, and (II) the potential at their common centre.

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) Work Done in Electric Field:**
The relation between electric field and potential is $dV = -\vec{E} \cdot d\vec{r}$.
$$V = -\int \vec{E} \cdot d\vec{r} = -\int (40x \hat{i}) \cdot (dx \hat{i} + dy \hat{j} + dz \hat{k})$$
$$V = -\int 40x \, dx = -20x^2$$
* Potential at start point A $(0, 3	ext{m})$: $V_A = -20(0)^2 = 0	ext{ V}$.
* Potential at end point B $(5	ext{m}, 0)$: $V_B = -20(5)^2 = -20(25) = -500	ext{ V}$.

The work done $W$ by an external agent to move a charge $q$ is $W_{ext} = q(V_B - V_A)$.
For a unit positive charge ($q = 1	ext{ C}$):
$$W = 1 	imes (-500 - 0) = \boxed{-500	ext{ J}}$$
*(Note: Work done by the electric field itself would be $+500	ext{ J}$. Standard convention for "work done in taking a charge" implies external work).*

**(ii) Concentric Spheres (Equal Surface Charge Density):**
Let $q$ be the charge on inner sphere (radius $r$) and $Q-q$ be the charge on outer sphere (radius $R$).
Since surface charge densities ($\sigma$) are equal:
$$\frac{q}{4\pi r^2} = \frac{Q-q}{4\pi R^2}$$
$$q R^2 = (Q-q) r^2 = Qr^2 - qr^2$$
$$q(R^2 + r^2) = Qr^2 \Rightarrow q = \frac{Qr^2}{R^2 + r^2}$$
The charge on the outer sphere is $Q-q = Q - \frac{Qr^2}{R^2 + r^2} = \frac{QR^2}{R^2 + r^2}$.

![Concentric Spheres Diagram](https://www.edvaya.com/images/cbse/physics/2025/set2/q32a-concentric-hollow-spheres-answer.webp)

**(I) Electric field at common centre:**
Inside any uniformly charged hollow sphere, the electric field is zero. The centre is inside both spheres.
Thus, $q_{enclosed} = 0 \Rightarrow \boxed{E = 0}$.

> **Alternatively:** $q_{en} = 0 \Rightarrow \phi_E = 0 \Rightarrow \oint \vec{E} \cdot d\vec{s} = 0 \Rightarrow E = 0$

**(II) Potential at common centre:**
The potential at the centre is the scalar sum of the potentials due to each shell at its surface.
$$V = V_{inner} + V_{outer} = \frac{k q}{r} + \frac{k (Q-q)}{R} \quad 	ext{where } k = \frac{1}{4\pi\epsilon_0}$$
Substitute the values of the charges:
$$V = \frac{k}{r} \left( \frac{Qr^2}{R^2 + r^2} ight) + \frac{k}{R} \left( \frac{QR^2}{R^2 + r^2} ight)$$
$$V = \frac{kQr}{R^2 + r^2} + \frac{kQR}{R^2 + r^2}$$
$$\boxed{V = \frac{kQ(r + R)}{r^2 + R^2}}$$

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