GATE 2025 PH – Question 49
A charge $q$ is placed a distance $d$ above an infinite grounded conducting plate at $z=0$. In $z>0$, $\phi=\phi_1+\phi_2$ with $$\phi_1=\frac q{4\pi\epsilon_0\sqrt{x^2+y^2+(z-d)^2}},\quad\phi_2=-\frac q{4\pi\epsilon_0\sqrt{x^2+y^2+(z+d)^2}}.$$ Which option(s) is/are correct?
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Correct answer: (D) $\phi_1$ satisfies Poisson’s equation in $z>0$
Explanation
**Method of images.** A charge $+q$ at height $d$ above a grounded plane is equivalent (for $z>0$) to the charge $+q$ and an image charge $-q$ at $z=-d$, as given by $\phi=\phi_1+\phi_2$.
- **A. Force magnitude $q/(16\pi\epsilon_0d^2)$.** The force on the charge is the Coulomb force from the image at distance $2d$:
$$F=\frac{q^2}{4\pi\epsilon_0(2d)^2}=\frac{q^2}{16\pi\epsilon_0d^2}.$$
The option has $q$ instead of $q^2$. ✗
- **B. The energy is $q^2/(8\pi\epsilon_0d)$.** The energy is $-\dfrac{q^2}{16\pi\epsilon_0d}$ (half of the interaction energy with the image, and negative). ✗
- **C. Surface density proportional to $1/\sqrt{x^2+y^2+d^2}$.** The induced charge density is $\sigma=-\dfrac{qd}{2\pi(x^2+y^2+d^2)^{3/2}}$, so it goes as $(x^2+y^2+d^2)^{-3/2}$. ✗
- **D. $\phi_1$ satisfies Poisson's equation in $z>0$.** True: it is the potential of the real point charge, which sits in $z>0$. ✓
Answer **D**.
Official GATE 2025 answer key: https://gate2026.iitg.ac.in/doc/download/2025_Key/PH_Keys.pdf