Two charged conducting spheres of radii a and b are connected by a wire. The ratio of the electric fields at their surfaces respectively, is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( b : a \)
Conductors joined by a wire come to the same potential. For a sphere, both the potential and the surface field depend on its charge and radius:
\( \displaystyle V = \frac{kQ}{R}, \qquad E = \frac{kQ}{R^{2}} = \frac{V}{R} \)
With the same \( V \) for both spheres, the surface field is inversely proportional to the radius:
\( \displaystyle \frac{E_a}{E_b} = \frac{V/a}{V/b} = \frac{b}{a} \)
So the smaller sphere has the stronger field at its surface.