Question

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

• A. potential at its surface is $150V$
• B. the electric field o the surface is $1500V/m$
• C. the charge on the sphere is ${\left(5/3\right)\times 10}^{-10}C$
• D. the electric potential at its centre is $225V$

Answer 1

Answer: (A), (B), (D)

Let radius of sphere is $R$ and charge on the sphere is $Q$
here, $V_1=k\frac{Q}{R+0.05}=100\Rightarrow kQ=100R+5...\left(1\right)$
and $V_2=k\frac{Q}{R+0.1}=75\Rightarrow kQ=75R+7.5...\left(2\right)$
Solving (1) and (2), $R=0.1m$ and $kQ=15$ where $k=\frac{1}{4\pi {ϵ}_0}\sim 9\times {10}^9$, Thus $Q=\left(15/9\right)\times {10}^{-9}=\left(5/3\right)\times {10}^{-9}C$
Potential at its surface is $V_R=\frac{kQ}{R}=\frac{15}{0.1}=150V$