Question

A metal sphere of radius $R$ cm is charged with $4\pi \mu$ C situated in air. If $\sigma$ is surface density of charge, $E$ is electric intensity at a distance $r$ from the centre of sphere, then $r$ is
($\varepsilon_0$ = permittivity of free space)

• A. $R\sqrt{\dfrac{\varepsilon_0 E}{\sigma}}$
• B. $R\sqrt{\dfrac{\sigma}{\varepsilon_0 E}}$
• C. $\sqrt{\dfrac{\varepsilon_0 E}{R\sigma}}$
• D. $\sqrt{\dfrac{R\sigma}{\varepsilon_0 E}}$

Hint:

Outside a charged conducting sphere, the electric field behaves as if all charge were concentrated at the centre.

Answer 1

The Correct Option is B

Step 1: Surface charge density of the sphere.
Total charge on the sphere is $Q$. Surface charge density:
\[ \sigma = \frac{Q}{4\pi R^2} \Rightarrow Q = 4\pi R^2 \sigma \]

Step 2: Electric field at distance $r$ from centre.
For $r \ge R$, electric field due to a charged sphere is:
\[ E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2} \]

Step 3: Substitute $Q$.
\[ E = \frac{1}{4\pi\varepsilon_0}\frac{4\pi R^2 \sigma}{r^2} = \frac{R^2 \sigma}{\varepsilon_0 r^2} \]

Step 4: Solve for $r$.
\[ r^2 = \frac{R^2 \sigma}{\varepsilon_0 E} \Rightarrow r = R\sqrt{\frac{\sigma}{\varepsilon_0 E}} \]

Step 5: Conclusion.
The distance from the centre is $r = R\sqrt{\dfrac{\sigma}{\varepsilon_0 E}}$.