A conducting sphere of radius a has charge Q | vedclass

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Question

Radius $=a$

charge $=Q$

radius $=2 a$ (inner)

outer radius $=3 a$

Electric field due to each sphere

$E_{1}=E_{3}=0$ (inside the inner c

Vedclass · Accepted Answer

$\frac{5}{{12}}\frac{{k{Q^2}}}{a}$

Vedclass · Answer

Radius $=a$

charge $=Q$

radius $=2 a$ (inner)

outer radius $=3 a$

Electric field due to each sphere

$E_{1}=E_{3}=0$ (inside the inner conductor from $O$ to a and from $2 a$ to $3 a$ in conducting shell)

$E_{2}$ and $E_{4}=\frac{K Q}{r^{2}}$

Now,

$d U=0.5 e\left\{E_{2} r^{2} \cdot d r(a \; to \; 2 a)\right\}+E_{4} 2 r^{2} d r\{3 a \; to\;  \alpha\}$

$=\frac{5}{12} \frac{Q^{2}}{4 \pi \epsilon_{0} a}$

$=\frac{5}{{12}}\frac{{k{Q^2}}}{a}$

A conducting sphere of radius a has charge $Q$ on it. It is enclosed by a neutral conducting concentric spherical shell having inner radius $2a$ and outer radius $3a.$ Find electrostatic energy of system.

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