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

Answer 1

Answer: (A), (C), (D)

potential at its surface is $150V$
the electric field o the surface is $1500V/m$
the electric potential at its centre is $225V$

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 $V_R=\frac{kQ}{R}=\frac{15}{0.1}=150V$
Field at its surface is $E_R=\frac{kQ}{R^2}=\frac{15}{{0.1}^2}=1500V/m$
Using Gauss's law field outside the sphere is
$E_{out}.4\pi r^2=\frac{Q}{{ϵ}_0}\Rightarrow E_{out}=\frac{kQ}{r^2}$
and inside is $E_{in}.4\pi r^2=\frac{\rho .\left(4/3\right)\pi r^3}{{ϵ}_0}$ where $Q=\rho .\left(4/3\right)\pi R^3$
thus, $E_{in}=\frac{kQr}{R^3}$
Potential at center is $V_C=-{\int }_{\infty }^{0}Edr=-{\int }_{\infty }^R{E}_{out}dr-{\int }_R^0{E}_{in}dr=-{\int }_{\infty }^0\frac{kQ}{r^2}-{\int }_R^0\frac{kQr}{R^3}dr=\frac{kQ}{R}+\frac{kQ}{2R}$

$=3/2V_R=3/2\times 150=225V$