WBJEE · Physics · Electrostatics
A sphere of radius \(R\) has a volume density of charge \(\rho=k r,\) where \(r\) is the distance from the centre of the sphere and \(k\) is constant. The magnitude of the electric field which exists at the surface of the sphere is given by \(\left(\varepsilon_{0}=\right.\) permittivity of free space)
- A \(\frac{4 \pi k R^{4}}{3 \varepsilon_{0}}\)
- B \(\frac{k R}{3 e_{0}}\)
- C \(\frac{4 \pi k R}{\varepsilon_{0}}\)
- D \(\frac{k R^{2}}{4 \varepsilon_{0}}\)
Answer & Solution
Correct Answer
(D) \(\frac{k R^{2}}{4 \varepsilon_{0}}\)
Step-by-step Solution
Detailed explanation
By Gauss's theorem…
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