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Question
The figure shows the field lines on a positive charge. Is the work done by the field in moving a small positive charge from Q to P positive or negative? Give reason.
Solution
Work done = q.(potential at Q − potential at P), where q is the small positive charge
The electric potential at a point distant r due to the field created by a positive charge Q is given by
\[V = \frac{1}{4 \pi\epsilon_0}\frac{q}{r}\]
\[ \because r_P < r_Q \]
\[ \Rightarrow V_P > V_Q\]
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