A charge q is kept at the centre of a sphere. What is the work done in moving a charge \( q_0 \) from one point on the surface of the sphere to the diametrically opposite point?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Zero
A point charge q gives the potential \( V=\frac{1}{4\pi\varepsilon_0}\frac{q}{r} \) at distance r. So the potential has the same value at every point of a sphere centred on the charge. Such a sphere is an equipotential surface. The work done in moving \( q_0 \) between two points is \( q_0 \) times the potential difference between them, whatever the path.
Both ends of the diameter are at the same distance from q, so \( V_2=V_1 \):
\( \begin{aligned} W &= q_0\,(V_2-V_1) \\ &= q_0\times0 \\ &= 0 \end{aligned} \)
NCERT (2026-27):
Equipotential Surfaces – “… For a single charge q, the potential is given by Eq. (2.8):
\( \displaystyle V=\frac{1}{4\pi\varepsilon_0}\,\frac{q}{r} \qquad (2.8) \)
This shows that V is a constant if r is constant. Thus, equipotential surfaces of a single point charge are concentric spherical surfaces centred at the charge.”
Equipotential Surfaces – “… the definition of an equipotential surface: there is no potential difference between any two points on the surface and no work is required to move a test charge on the surface.”