Question
Write down the equation of electric potential for a point $-$ outside the spherical shell $-$ on the surface of a spherical shell and $-$ inside the spherical shell.
✓
Answer
$\rightarrow $ Generally, for a spherical shell, the electric field for a point outside the surface of shell is such that the entire charge of shell can be considered as concentrated at its centre.
$\rightarrow $ The value of electric potential that we get for a point outside the surface of shell is also got $($derived$)$ in a similar way.
$V =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}(r \geq R )$
Where, $q -$ total electric charge on the shell
$R -$ radius of the shell
$r -$ the distance of the point from the centre of the shell.
$\rightarrow $ The electric field inside the shell is zero, therefore potential inside remains constant.
$\rightarrow $ Because as there is no electric field inside, the work done in taking the charge from one point to the other is zero.
$\therefore \text { In } W =q \Delta V ,$
$\text { because } W =0$
$\therefore \Delta V =0$
$\therefore V =\text { constant }$
and that value of potential is same as the value on the surface of shell.
$\therefore V =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{ R } \quad(\because r \leq R )$
$\rightarrow $ Below figure $($graph$)$ shows the variation of electric potential $( V )$ versus distance from the
$\rightarrow $ The value of electric potential that we get for a point outside the surface of shell is also got $($derived$)$ in a similar way.
$V =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{r}(r \geq R )$
Where, $q -$ total electric charge on the shell
$R -$ radius of the shell
$r -$ the distance of the point from the centre of the shell.
$\rightarrow $ The electric field inside the shell is zero, therefore potential inside remains constant.
$\rightarrow $ Because as there is no electric field inside, the work done in taking the charge from one point to the other is zero.
$\therefore \text { In } W =q \Delta V ,$
$\text { because } W =0$
$\therefore \Delta V =0$
$\therefore V =\text { constant }$
and that value of potential is same as the value on the surface of shell.
$\therefore V =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{q}{ R } \quad(\because r \leq R )$
$\rightarrow $ Below figure $($graph$)$ shows the variation of electric potential $( V )$ versus distance from the
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