GATE 2024 EE Q.51: solution and answer

Official answer: -1.7 to -1.5Source: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Field of a point charge (Coulomb's law): $\mathbf{E}=\frac{q}{4\pi {ϵ}_0}\frac{\mathbf{R}}{|\mathbf{R}{|}^3}$, where $\mathbf{R}$ points from the charge to the field point $P(0,0.5,0)$. Write $k=\frac{1}{4\pi {ϵ}_0}$, with ${ϵ}_0=8.854\times 10^{-12}\text{F/m}$. $k$ cancels at the end, so its value is never needed.

2. Charge $Q$ at $(-1,0,0)$: ${\mathbf{R}}_1={\hat{a}}_x+0.5{\hat{a}}_y$ and $|{\mathbf{R}}_1|=\sqrt{1.25}$. Charge $Q$ at $(1,0,0)$: ${\mathbf{R}}_2=-{\hat{a}}_x+0.5{\hat{a}}_y$ and $|{\mathbf{R}}_2|=\sqrt{1.25}$.

3. By symmetry the $x$-components cancel and the $y$-components add: $${\mathbf{E}}_1+{\mathbf{E}}_2=kQ\frac{2(0.5)}{(1.25{)}^{3/2}}{\hat{a}}_y=\frac{kQ}{(1.25{)}^{3/2}}{\hat{a}}_y\approx 0.7155kQ{\hat{a}}_y$$

4. Charge $\alpha Q$ at $(0,-1,0)$: ${\mathbf{R}}_3=1.5{\hat{a}}_y$, so $${\mathbf{E}}_3=\frac{k\alpha Q}{(1.5{)}^2}{\hat{a}}_y=\frac{\alpha kQ}{2.25}{\hat{a}}_y\approx 0.4444\alpha kQ{\hat{a}}_y$$

5. Superposition, with the total field set to zero (divide through by $kQ$): $$\frac{1}{(1.25{)}^{3/2}}+\frac{\alpha }{2.25}=0\Rightarrow \alpha =-\frac{2.25}{(1.25{)}^{3/2}}$$

6. Exact value: $(1.25{)}^{3/2}={\left(\frac{5}{4}\right)}^{3/2}=\frac{5\sqrt{5}}{8}$, so $\alpha =-\frac{9}{4}\cdot \frac{8}{5\sqrt{5}}=-\frac{18}{5\sqrt{5}}=-\frac{18\sqrt{5}}{25}\approx -1.610$.

7. Sign check: both $Q$ charges push $P$ in the $+y$ direction. The third charge sits below $P$, so it has to pull $P$ downward, toward itself, and that needs $\alpha <0$. Rounded to 1 decimal place, $\alpha =-1.6$.