A)
B)
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D)
Correct Answer: B
Solution :
Outside the spherical charge, the intensity of electric field at a point P situated at a distance r from the centre of the charge is \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\,\frac{q}{{{r}^{2}}}\] (if r > R) On the surface of spherical charge the electric field is given by \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\,\frac{q}{{{r}^{2}}}\] (if r=R) and inside the spherical charge the electric field: is. \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\,\frac{qr}{{{R}^{3}}}\] Hence, the variation of E-r is a follows: Note: At the centre of sphere, r = 0, so E = 0.You need to login to perform this action.
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