
A solid hemispherical uniform charged body having charge $Q$ is kept symmetrically along the $y$ -axis as…

- $\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{d}$
- less than $\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{d}$
- more than $\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{d}$ and less than $\frac{2}{4 \pi \varepsilon_{0}} \frac{Q}{d}$
- more than $\frac{2}{4 \pi \varepsilon_{0}} \frac{Q}{d}$
Solution
So potential at point $P$ due to this spherical charge $=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 Q}{d}$
Hence potential due to hemisphere $=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{d}$
$\therefore \quad$ (a)
Asked in: JEE Mains - Electrostatics - Test 3