In the figure shown, the electric potential energy of the system is (q is at the centre of the conducting neutral spherical shell of inner radius a and outer radius b )

Question

In the figure shown, the electric potential energy of the system is (q is at the centre of the conducting neutral spherical shell of inner radius a and outer radius b ) (A) 0 (B) (k q2/2 b) (C) (k q2/2 b)-(k q2/2 a) (D) (k q2/2 a)-(k q2/2 b)

Tardigrade Admin · Accepted Answer

U=(-k q q/a)+(k q q/b)-(k q q/b)+(k q2/2 a)+(k q2/2 b) =(k q2/2 b)-(k q2/2 a) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/a746da4acee1f21f1602cfbe66a31bcd-.png" />

Q. In the figure shown, the electric potential energy of the system is $(q$ is at the centre of the conducting neutral spherical shell of inner radius $a$ and outer radius $b$ )

0

$\frac{k q ^{2}}{2 b}$

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

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

$U = \frac{- k qq}{a} + \frac{k qq}{b} - \frac{k qq}{b} + \frac{k q ^{2}}{2 a} + \frac{k q ^{2}}{2 b}$
$= \frac{k q ^{2}}{2 b} - \frac{k q ^{2}}{2 a}$

Q. In the figure shown, the electric potential energy of the system is (q is at the centre of the conducting neutral spherical shell of inner radius a and outer radius b )

0

2bkq2​

2bkq2​−2akq2​

2akq2​−2bkq2​

U=a−kqq​+bkqq​−bkqq​+2akq2​+2bkq2​ =2bkq2​−2akq2​

Q. In the figure shown, the electric potential energy of the system is $(q$ is at the centre of the conducting neutral spherical shell of inner radius $a$ and outer radius $b$ )

$\frac{k q ^{2}}{2 b}$

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

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

$U = \frac{- k qq}{a} + \frac{k qq}{b} - \frac{k qq}{b} + \frac{k q ^{2}}{2 a} + \frac{k q ^{2}}{2 b}$
$= \frac{k q ^{2}}{2 b} - \frac{k q ^{2}}{2 a}$