GATE 2018 EE Q.10: solution and answer

Official answer: DSource: Official answer key, GATE 2018 EE (IIT Guwahati)

1. Method of images: replace the grounded plane $z=0$ with an image charge $Q^{\prime }=-1\text{nC}$ at $(0,0,-0.2)$. The pair keeps $V=0$ everywhere on the plane, so the field in the region $z>0$ is unchanged. The field point $(0,0,0.1)$ lies in that region.

2. Coulomb constant, with ${ϵ}_0=8.85\times 10^{-12}\text{F/m}$: $$\frac{1}{4\pi {ϵ}_0}=\frac{1}{4\pi (8.85\times 10^{-12})}\approx 8.992\times 10^9\text{m/F}$$

3. Real charge $+1\text{nC}$ at $z=0.2$: the field point is $0.1\text{m}$ below it. The field of a positive charge points away from the charge, so here it points in $-{\hat{a}}_z$: $$E_1=\frac{8.992\times 10^9\times 10^{-9}}{(0.1{)}^2}=899.18\text{V/m},{\mathbf{E}}_1=-899.18{\hat{a}}_z\text{V/m}$$

4. Image charge $-1\text{nC}$ at $z=-0.2$: the field point is $0.3\text{m}$ above it. The field of a negative charge points toward the charge, so this one also points in $-{\hat{a}}_z$: $$E_2=\frac{8.992\times 10^9\times 10^{-9}}{(0.3{)}^2}=99.91\text{V/m},{\mathbf{E}}_2=-99.91{\hat{a}}_z\text{V/m}$$

5. Superposition: $$E_z=-(899.18+99.91)\approx -999.09\text{V/m}$$ which is option (D).