Question

A solid metallic sphere has a charge +3Q. Concentric with this sphere is a conducting spherical shell having charge -Q. The radius of the sphere is 'A' and that of the spherical shell is 'B' ($B>A$). The electric field at a distance 'R' ($A<R<B$) from the centre is ($\epsilon_{0} =$ permittivity of vacuum)

• A. $\frac{Q}{2\pi\epsilon_{0}R}$
• B. $\frac{3Q}{2\pi\epsilon_{0}R}$
• C. $\frac{3Q}{4\pi\epsilon_{0}R^{2$
• D. $\frac{4Q}{2\pi\epsilon_{0}R^{2$

Hint:

Logic Tip: Remember that the electric field inside a conducting shell due to the shell's own charge is always zero.

Answer 1

The Correct Option is C

Concept:
According to Gauss's Law, the electric field at a point depends only on the charge enclosed by the Gaussian surface passing through that point.
Step 1: Identify the enclosed charge.
For a distance $R$ where $A<R<B$, the Gaussian surface (a sphere of radius $R$) encloses only the inner solid metallic sphere. The charge on the outer shell does not contribute to the field inside its own radius. $$\text{Enclosed Charge } q = +3Q \text{ }$$
Step 2: Apply the electric field formula.
The electric field $E$ due to a point charge or a spherical distribution at distance $R$ is: $$E = \frac{1}{4\pi\epsilon_{0\frac{q}{R^{2 \text{ }$$ $$E = \frac{1}{4\pi\epsilon_{0\frac{(3Q)}{R^{2 = \frac{3Q}{4\pi\epsilon_{0}R^{2 \text{ }$$