GATE 2025 EE Q.25: solution and answer

Official answer: CSource: Official answer key, GATE 2025 EE (IIT Roorkee)

1. Symmetry. The charge is spherically symmetric, so $\mathbf{E}=E_R{\hat{a}}_R$ and $E_R$ depends only on $R$. Gauss's law over a concentric sphere of radius $R$ gives $E_R(4\pi R^2)=Q_{enc}/{\epsilon }_0$, with ${\epsilon }_0\approx 8.854\times 10^{-12}\text{F/m}$.

2. Inside ($0\le R\le a$). The enclosed charge is $Q_{enc}=(-3{\rho }_0)\cdot \frac{4}{3}\pi R^3=-4\pi {\rho }_0{R}^3$, so $$E_R=\frac{-4\pi {\rho }_0{R}^3}{4\pi {\epsilon }_0{R}^2}=-\frac{{\rho }_{0}R}{{\epsilon }_0}$$ This is a straight line through the origin with a negative slope.

3. At the surface ($R=a$). $E_R(a)=-\frac{{\rho }_{0}a}{{\epsilon }_0}$. This is the most negative value the field reaches.

4. Outside ($R>a$). Now the sphere encloses all the charge, $Q=-4\pi {\rho }_0{a}^3$, so $$E_R=\frac{-4\pi {\rho }_0{a}^3}{4\pi {\epsilon }_0{R}^2}=-\frac{{\rho }_0{a}^3}{{\epsilon }_0{R}^2}$$ This is negative and rises towards zero as $1/R^2$. At $R=a$ it equals $-{\rho }_{0}a/{\epsilon }_0$, the same as the inside value, so the field is continuous at the surface.

5. Pick the figure. The field is negative everywhere. It falls linearly to $-{\rho }_{0}a/{\epsilon }_0$ at $R=a$, then returns to zero as $1/R^2$. That is Fig. (iii), option (C). Figs. (i) and (ii) have the wrong sign: they would describe positive charge. Fig. (iv) has the right shape but twice the magnitude, which would need $\rho =-6{\rho }_0$.