Question

At the distance of 5 cm and 10 cm from surface of a uniformly charged solid sphere, the potentials are 100 V and 75 V respectively. Then

• A. Potential at its surface is 150 V
• B. The charge on the sphere is $\frac{50}{3}\times {10}^{-10}C$
• C. The electric field on the surface is 1500 V/m
• D. The electric potential at its centre is 25 V

Answer 1

Answer: (A), (B), (C)

Potential at its surface is 150 V
The charge on the sphere is $\frac{50}{3}\times {10}^{-10}C$
The electric field on the surface is 1500 V/m

Let $R=$ radius of sphere and $q=$ charge on sphere.

here, $100=\frac{kq}{R+0.05}\Rightarrow 100R+5=kq...\left(1\right)$ and

$75=\frac{kq}{R+0.1}\Rightarrow 75R+7.5=kq...\left(2\right)$

from (1) and (2), $75R+7.5=100R+5$

or $25R=2.5\Rightarrow R=0.1m$

using (1), $kq=100\left(0.1\right)+5=10+5=15$

$V_{surface}=\frac{kq}{R}=15/0.1=150V$

Charge, $q=15/k=\frac{15}{9\times {10}^9}=50/3\times {10}^{-10}C$ where $k=1/4\pi {ϵ}_0$

$E_{surface}=\frac{kq}{R^2}=\frac{15}{(0.1{)}^2}=1500V/m$