Question

At a distance of $5cm$ and $10cm$ outward 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 charge on the sphere is $\left(\frac{5}{3}\right)\times {10}^{10}C$
• C. the electric field on the surface is $1500V{m}^{-1}$
  
• D. the electric potential at its center is $0V$

Answer 1

Answer: (A), (C)

the electric field on the surface is $1500V{m}^{-1}$


We know that,
$V_0=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{R}$
$V=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{(R+d)}$
$R$ is radius, and $d$ is distance from surface
$100=\frac{KQ}{(R+5)\times {10}^{-2}}.......\left(i\right)$
$75=\frac{KQ}{(R+10)\times {10}^{-2}}.......\left(ii\right)$
From (i) and (ii), $R=10cm$ and
$Q=\frac{50}{3}\times {10}^{-10}C$
Potential at surface is
$V_0=\frac{KQ}{R}=\frac{9\times {10}^9\times 50}{10\times {10}^{-2}\times 3}\times {10}^{-10}C=150V$
Electric field on surface is
$E_0=\frac{KQ}{R^2}=1500V{m}^{-1}$
Potential at the center is $V_c=\frac{3}{2}{V}_0=225V$