Part (a)(i): Work Done in an Electric Field

Step 1 (Formula / Governing Law & Given Data): [1 Mark]

The work done $W$ in moving a charge $q_0$ in an electric field $\vec{E}$ is given by the line integral:

$$W = -\int_{A}^{B} q_0 \vec{E} \cdot d\vec{r}$$

Given: $\vec{E} = 40x\hat{i}$, $q_0 = 1\text{ C}$, Point $A = (0, 3)$, Point $B = (5, 0)$.

Step 2 (Step-by-Step Mathematical Working): [1 Mark]

Since $\vec{E}$ only has an $x$-component, the work done depends only on the change in the $x$-coordinate:

$$W = -q_0 \int_{x_i}^{x_f} E_x dx = -1 \int_{0}^{5} 40x \, dx$$ $$W = -40 \left[ \frac{x^2}{2} \right]_{0}^{5} = -40 \left[ \frac{25}{2} - 0 \right]$$ $$W = -40 \times 12.5$$

Step 3 (Final Concluded Answer): [0.5 Mark]

The work done is $-500\text{ J}$.

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Part (a)(ii): Concentric Hollow Spheres

Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

Given surface charge densities are equal: $\sigma_r = \sigma_R = \sigma$.
Total charge $Q = q_r + q_R = \sigma(4\pi r^2) + \sigma(4\pi R^2) = 4\pi\sigma(r^2 + R^2)$.
Therefore, $\sigma = \frac{Q}{4\pi(r^2 + R^2)}$.

Step 2 (Step-by-Step Mathematical Working): [1.5 Mark]

(I) Electric Field at the centre: By the shell theorem, the electric field inside any uniformly charged spherical shell is zero. Since the point is inside both shells, the net field is:

$$E_{net} = 0 + 0 = 0$$

(II) Potential at the centre: The potential at the centre is the sum of potentials due to both shells:

$$V = \frac{1}{4\pi\epsilon_0} \left( \frac{q_r}{r} + \frac{q_R}{R} \right)$$ $$V = \frac{1}{4\pi\epsilon_0} \left( \frac{\sigma 4\pi r^2}{r} + \frac{\sigma 4\pi R^2}{R} \right) = \frac{\sigma}{\epsilon_0} (r + R)$$

Step 3 (Final Concluded Answer): [0.5 Mark]

Substituting $\sigma$: $$V = \frac{Q}{4\pi\epsilon_0(r^2 + R^2)} (r + R)$$

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💡 EL10 Concept Insight:

Think of the electric field inside a hollow ball like a perfectly balanced room where every wall pulls you equally in opposite directions