Question

At distance 5cm and 10cm outwards from the surface of a uniformly charged solid sphere, the potentials are 100V and 75V respectively.
A. potential at its surface is 180V
B. the charge on the sphere is (5/3) x 10"C
C. the electric field on the surface is 1500 V /m
D. the electric potential at its centre is 225V

Answer 1

Hint:We know that the potential outside the uniformly charged solid sphere is given as $V = \dfrac{{kq}}{r}$where k is the electric force constant, q is the charge on the sphere and r is the distance of the point from the centre of the sphere. Using the equation we can find the charge on the sphere. According to the gauss's law, $EA = \dfrac{{{q_{enclosed}}}}{{{\varepsilon _0}}}$. Electric field(E) outside, inside and on the surface of the sphere can be found out. Potential at the centre of the sphere can be found using the relation.

Complete step by step answer:
The potential outside the solid sphere is given as $V = \dfrac{{kq}}{r}$ where k is the electric force constant, q is the charge on the sphere and r is the distance of the point from the centre of the sphere. Given at a distance 5 cm from the surface the potential is 100V and at a distance of 10cm from the surface the potential is 75 V.
Substituting these values we get,
$100 = \dfrac{{kq}}{{R + 0.05}}$
$\Rightarrow 100R + 5 = kq$ --------(1)
$\Rightarrow 75 = \dfrac{{kq}}{{R + 0.1}}$
$\Rightarrow 75R + 7.5 = kq$ --------(2)
Subtracting equation 2 from 1 .
$ 25R - 2.5 = 0 \\
\Rightarrow R = 0.1m \\ $
We get the value of radius R=0.1m.
Substituting the value of R and k$ = 9 \times {10^9}$in equation (1),
$100 \times 0.1 + 5 = 9 \times 1{0^9}q \\
\Rightarrow q = \dfrac{{15}}{{9 \times {{10}^9}}} \\
\Rightarrow q= \left( {\dfrac{5}{3}} \right) \times {10^{ - 9}}C \\ $
Potential at the surface is $ = \dfrac{{kq}}{R} = \dfrac{{9 \times {{10}^9} \times \left( {\dfrac{5}{3}} \right) \times {{10}^{ - 9}}}}{{0.1}} = 150V$
Electric field at the surface is
$E = \dfrac{{kq}}{{{R^2}}}\\
\Rightarrow E = \dfrac{{9 \times 1{0^9} \times \left( {\dfrac{5}{3}} \right) \times {{10}^{ - 9}}}}{{{{0.1}^2}}}\\
\Rightarrow E = 1500V/m$
According to gauss's law the electric field through a surface is directly proportional to the charge enclosed. The electric field outside the sphere is,
\[{E_{out}}.4\pi {r^2} = \dfrac{q}{{{ \in _0}}} \\
\Rightarrow {E_{out}} = \dfrac{{kq}}{{{r^2}}}\]
The electric field inside the sphere is ${E_{in}} = \dfrac{{kqr}}{{{R^2}}}$.
The potential at the centre can be given as,
${V_C} = - \int\limits_\infty ^0 {Edr = - \int\limits_\infty ^R {{E_{out}}dr} } - \int\limits_R^0 {{E_{in}}dr}$
$\Rightarrow{V_C} = - \int\limits_\infty ^R {\dfrac{{kq}}{{{r^2}}}dr} - \int\limits_R^0 {\dfrac{{kqr}}{{{R^2}}}} dr $
$\Rightarrow{V_C} = \dfrac{{kq}}{R} + \dfrac{{kq}}{{2R}}\\
\Rightarrow{V_C} = \dfrac{3}{2}\dfrac{{kq}}{R}$
$\Rightarrow {V_C} = \dfrac{3}{2}{V_{Surface}}\\
\Rightarrow {V_C} = \dfrac{3}{2} \times 150 \\
\therefore {V_C} = 225V$

Hence,option D is the correct answer.

Note:One should be aware of the gauss's law which states that the electric field in an enclosed surface is directly proportional to the charge enclosed by that surface (this helps us to find the electric field at any surface). The most important point is to remember the relation between electric field E and electric potential V $\left( {V = - \int {Edr} } \right)$to solve such problems.