Question

A solid conducting sphere of radius $2R$, carrying charge $Q$ is surrounded by two point charges $Q$ and $2Q$ as shown in the figure. The electric field at point $P$ due to the induced charge on conducting sphere is

• A. $\frac{7}{16}\frac{KQ}{R^2}$ towards right
• B. $\frac{1}{8}\frac{KQ}{R^2}$ towards right
• C. $\frac{KQ}{R^2}$ towards right
• D. Zero

Answer 1

The correct option is A $\frac{7}{16}\frac{KQ}{R^2}$ towards right

Total distance covered $=3/R+3/R+1/R=7/R$

$2Q$ and charge $={\left(2Q\right)}^{2}Q$

$=24/Q$

momentum $R\times \left(4/Q\right)=16/Q\times R$

Hence,

Electric field $=\frac{7\times Q}{16R^2}K$

$=\frac{7}{16}\times \frac{QK}{R^2}$ towards right.