Question 863: Electrostats & Capacitance

Question

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 '$$B$$'. $$(B > A)$$. The electric field at a distance '$$\mathrm{R}$$' $$(\mathrm{A} < \mathrm{R} < \mathrm{B})$$ from the centre is ( $$\varepsilon_0=$$ permittivity of vacuum)

Accepted Answer

We know the electric field inside a shell is zero. So, we only consider the electric field due to the solid sphere.
$$\begin{aligned}
& 	herefore \quad 	ext { Using } E=\frac{1}{4 \pi \varepsilon_0} \frac{q}{r^2} 	ext {, we get } \
& \quad E=\frac{1}{4 \pi \varepsilon_0} \frac{(3 Q)}{R^2}
\end{aligned}$$

Answer

$$\frac{\mathrm{Q}}{2 \pi \varepsilon_0 \mathrm{R}}$$

Answer

$$\frac{3 Q}{2 \pi \varepsilon_0 R}$$

Answer

$$\frac{3 \mathrm{Q}}{4 \pi \varepsilon_0 \mathrm{R}^2}$$

Answer

$$\frac{4 Q}{2 \pi \varepsilon_0 R^2}$$

Electrostats & Capacitance

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