Q. 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 $b\ (b > a).$ What is the electric field at a distance $R(a < R < b)$ from the centre?
A) $\frac{4\mathcal{Q}}{2\pi\varepsilon_{0}R^{2}}$
B) $\frac{3\mathcal{Q}}{4\pi\varepsilon_{0}R^{2}}$
C) $\frac{3\mathcal{Q}}{2\pi\varepsilon_{0}R^{2}}$
D) $\frac{\mathcal{Q}}{2\pi\varepsilon_{0}R}$
Explanation
(b) The electric field inside a spherical charge is everywhere zero, that is
$E = 0$
But point P is outside the inner sphere, hence for a point very close to the surface the intensity of electric field is given by
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$E = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{R^{2}}$
Given, $q = + 3\mathcal{Q}$
Therefore,
$E = \frac{1}{4\pi\varepsilon_{0}}\frac{3\mathcal{Q}}{R^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.