Question

A charge $+ q$ is placed at the origin O of x – y axes as shown in the figure. The work done in taking a charge Q from A to B along the straight line AB is.

• A. $\frac{qQ}{4\pi\varepsilon_{0}}\left( \frac{a - b}{ab} \right)$
• B. $\frac{qQ}{4\pi\varepsilon_{0}}\left( \frac{b - a}{ab} \right)$
• C. $\frac{qQ}{4\pi\varepsilon_{0}}\left( \frac{b}{a^{2}} - \frac{1}{b} \right)$
• D. $\frac{qQ}{4\pi\varepsilon_{0}}\left( \frac{b}{b^{2}} - \frac{1}{b} \right)$

Answer 1

Answer: (B)

$\frac{qQ}{4\pi\varepsilon_{0}}\left( \frac{b - a}{ab} \right)$

Answer 2

: Potential at point A is,

$V_{A} = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{a}$

Potential at pint B is,

$V_{B} = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{b}$

Work done in taking a charge Q from A to B is,

$W = Q(V_{B} - V_{A}) = \frac{Qq}{4\pi\varepsilon_{0}}\left\lbrack \frac{1}{b} - \frac{1}{a} \right\rbrack = \frac{Qq}{4\pi\varepsilon_{0}}\left\lbrack \frac{b - a}{ab} \right\rbrack$