QuestionElectrostatics / conductors
A spherical conductor of radius \(a\) has a cavity of radius \(b\) at its centre such that \(a > b\). A positive charge \(q\) is placed at its centre (see in the figure). The electric fields in the regions \(b < r < a\) and \(r > a\) are respectively

- (a) \(\frac{q}{4\pi\varepsilon_0 a^2}\) inward and \(\frac{q}{4\pi\varepsilon_0 b^2}\) outward
- (b) \(\frac{q}{4\pi\varepsilon_0 r^2}\) inward and zero
- (c) zero and \(\frac{q}{4\pi\varepsilon_0 r^2}\) outward
- (d) zero and zero
Part of VKC Basic
Read the answer and its explanation with VKC Basic
VKC Basic (₹999 for 6 months) opens every solved answer, the notes, the handbook and the daily answer archive. Every Mains test series includes it.