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.

Answer 1

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

\[V = \frac{1}{4 \pi\epsilon_0}\frac{q}{r}\]

\[ \because r_P < r_Q \]

\[ \Rightarrow V_P > V_Q\]

∴ Work done will be negative.