As per the diagram a point charge $+q$ is placed at the origin $\mathrm{O}$. Work done in taking another…

As per the diagram a point charge $+q$ is placed at the origin $\mathrm{O}$. Work done in taking another point charge $-\mathrm{Q}$ from the point $\mathrm{A}$ [coordinate $(0, a)$ ] to another point $\mathrm{B}$ [coordinates $(a, 0)$ ] along the straight path $\mathrm{AB}$ is:

1. zero
2. $\left(\frac{q Q}{4 \pi \varepsilon_0} \frac{1}{a^2}\right) \cdot \sqrt{2} a$
3. $\left(\frac{-q Q}{4 \pi \varepsilon_0} \frac{1}{a^2}\right) \cdot \sqrt{2} a$
4. $\left(\frac{q Q}{4 \pi \varepsilon_0} \frac{1}{a^2}\right) \cdot \frac{a}{\sqrt{2}}$

Solution

Work done is equal to zero because the potential of A and B are the same $=\frac{1}{4 \pi \epsilon_0} \frac{q}{a}$ No work done if a particle does not change its potential energy. i.e. initial potential energy $=$ final potential energy.

Asked in: NEET 2005

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