JEE Advance - Physics (2007 - Paper 2 Offline - No. 7)
Positive and negative point charges of equal magnitude are kept at $$\left(0,0, \frac{a}{2}\right)$$ and $$\left(0,0, \frac{-a}{2}\right)$$, respectively. The work done by the electric field when another positive point charge is moved from $$(-a, 0,0)$$ to $$(0, a, 0)$$ is
positive
negative
zero
depends on the path connecting the initial and final positions
Explanation
Consider that,
$$\mathrm{P}=(-a, 0,0)$$ and $$\mathrm{Q}=(0, a, 0)$$ points lie on the equatorial plane of dipole.
The charges makes an electric dipole.
Therefore, potential at $$\mathrm{P}=$$ Potential at $$\mathrm{Q}=0$$
Work done $$(\mathrm{W})=q\left(\mathrm{~V}_{\mathrm{P}}-\mathrm{V}_{\mathrm{Q}}\right)=0$$
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