A point charge Q is placed at the centre of a spherical cavity of radius b carved inside a solid conducting sphere of radius a. Then total energy of the system is [k=(1/4 π ε0)]

Question

A point charge Q is placed at the centre of a spherical cavity of radius b carved inside a solid conducting sphere of radius a. Then total energy of the system is [k=(1/4 π ε0)] (A) (k Q2/2 a)-(k Q2/2 b) (B) (k Q2/2 a)+(k Q2/2 b) (C) (k Q2/a)+(k Q2/b) (D) (k Q2/a)-(k Q2/b)

Tardigrade Admin · Accepted Answer

-Q and Q will be induced uniformly as shown. Total energy = Self-energy + interaction energy <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/2c8a7edb6f4b8ea0ca0672ae47924d37-.png" /> = (k(-Q)2/2 b)+(k(Q)2/2 a)+(k(Q)(-Q)/b)+(k(Q)(+Q)/a) +(k(Q)(-Q)/a) =k Q2((1/2 a)-(1/2 b))

Q. A point charge $Q$ is placed at the centre of a spherical cavity of radius $b$ carved inside a solid conducting sphere of radius $a$ . Then total energy of the system is $[k = \frac{1}{4 π ε _{0}}]$

$\frac{k Q ^{2}}{2 a} - \frac{k Q ^{2}}{2 b}$

$\frac{k Q ^{2}}{2 a} + \frac{k Q ^{2}}{2 b}$

$\frac{k Q ^{2}}{a} + \frac{k Q ^{2}}{b}$

$\frac{k Q ^{2}}{a} - \frac{k Q ^{2}}{b}$

$- Q$ and $Q$ will be induced uniformly as shown.
Total energy $=$ Self-energy $+$ interaction energy

$= \frac{k ( - Q ) ^{2}}{2 b} + \frac{k ( Q ) ^{2}}{2 a} + \frac{k ( Q ) ( - Q )}{b} + \frac{k ( Q ) ( + Q )}{a}$
$+ \frac{k ( Q ) ( - Q )}{a}$
$= k Q^{2} (\frac{1}{2 a} - \frac{1}{2 b})$

Q. A point charge Q is placed at the centre of a spherical cavity of radius b carved inside a solid conducting sphere of radius a. Then total energy of the system is [k=4πε0​1​]

2akQ2​−2bkQ2​

2akQ2​+2bkQ2​

akQ2​+bkQ2​

akQ2​−bkQ2​