WBJEE · Physics · Electrostatics
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 \(A B\) is 
- A \(\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{a-b}{a b}\right)\)
- B \(\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{b-a}{a b}\right)\)
- C \(\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{b}{a^{2}}-\frac{1}{b}\right)\)
- D \(\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{a}{b^{2}}-\frac{1}{b}\right)\)
Answer & Solution
Correct Answer
(A) \(\frac{q Q}{4 \pi \varepsilon_{0}}\left(\frac{a-b}{a b}\right)\)
Step-by-step Solution
Detailed explanation
We know that and \(\quad V=\frac{Q}{4 \pi \varepsilon_{0} r}\) \(\begin{aligned} \text { Hence, } \quad W &=q\left[\frac{Q}{4 \pi \varepsilon_{0} b}-\frac{Q}{4 \pi \varepsilon_{0} a}\right] \\ &=\frac{Q q}{4 \pi \varepsilon_{0}}\left[\frac{a-b}{a b}\right] \end{aligned}\)
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