The surface charge density of a sphere of radius r is $σ$ . It is placed at a point A. The electric field intensity at B due to this sphere is E. Another charged sphere of radius 2r is placed at B. If the intensity at the centre of line joining A and B is E/2 , then the surface charge density of B is ( separation between the points A and B is much much greater than r )

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{- 7 σ}{32}$

$\frac{7 σ}{32}$

$\frac{σ}{2}$

$- \frac{σ}{2}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

The field at B due to A, $E = \frac{1}{4 {πε}_{0}} \frac{q}{d^{2}}$ $= \frac{1}{4 {πε}_{0}} \frac{σ (4 {πr}^{2})}{d^{2}}$

The field at P is $\frac{E}{2}$ which is equal to sum of fields due to A & B.

$\frac{E}{2} = E_{1} + E_{2}$

$\begin{array}{l} \Rightarrow \frac{E}{2} = \frac{1}{4 {πε}_{0}} (\frac{4 {πr}^{2} σ}{\frac{d^{2}}{4}}) - \frac{1}{4 {πε}_{0}} (\frac{4 π (2 r)^{2} σ'}{\frac{d^{2}}{4}}) \\ \Rightarrow \frac{1}{2} [\frac{σ 4 {πr}^{2}}{4 {πε}_{0} d^{2}}] = \frac{16 {πr}^{2} σ}{4 {πε}_{0} d^{2}} - \frac{64 {πr}^{2} σ'}{4 {πε}_{0} d^{2}}; \end{array} ∴ σ' = \frac{7 σ}{32}$