Q147 / 150 · MHT CET 2023 (09 May Shift 1) · Physics

A solid metallic sphere has a charge $+3 Q$. Concentric with this sphere is a conducting spherical shell having charge $-\mathrm{Q}$. The radius of the sphere is ' $A$ ' and that of the spherical shell is ' $\mathrm{B}$ '. $(\mathrm{B}>\mathrm{A})$. The electric field at a distance ' $R$ ' ( $A < R < B)$ from the centre is ( $\varepsilon_0=$ permittivity of vacuum)

  • A
    $\frac{\mathrm{Q}}{2 \pi \varepsilon_0 \mathrm{R}}$
  • B
    $\frac{\mathrm{3Q}}{2 \pi \varepsilon_0 \mathrm{R}}$
  • Correct answer
    C
    $\frac{\mathrm{3Q}}{4 \pi \varepsilon_0 \mathrm{R^2}}$
  • D
    $\frac{\mathrm{4Q}}{2 \pi \varepsilon_0 \mathrm{R^2}}$

Solution

$\begin{aligned} \therefore \quad & \text { Using } \mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{\mathrm{r}^2}, \text { we get } \\ & \mathrm{E}=\frac{1}{4 \pi \varepsilon_0} \frac{(3 \mathrm{Q})}{\mathrm{R}^2}\end{aligned}$

Hence, the correct answer is option: 'C'.

Relevant Questions