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Question: Answered & Verified by Expert
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:

PhysicsElectrostaticsNEETNEET 2005
Options:
  • A zero
  • B $\left(\frac{q Q}{4 \pi \varepsilon_0} \frac{1}{a^2}\right) \cdot \sqrt{2} a$
  • C $\left(\frac{-q Q}{4 \pi \varepsilon_0} \frac{1}{a^2}\right) \cdot \sqrt{2} a$
  • D $\left(\frac{q Q}{4 \pi \varepsilon_0} \frac{1}{a^2}\right) \cdot \frac{a}{\sqrt{2}}$
Solution:
1688 Upvotes Verified Answer
The correct answer is: zero
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.

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