12.6 Potential Due To Uniformly Charged Spherical Shell

  Electrostatics

    1.0 Introduction
    2.0 Electric charge
    3.0 Coulomb's law
    4.0 Principle of superposition
    5.0 Continuous charge distribution
    6.0 Electric field
    7.0 Electric field lines
    8.0 Insulators and conductors
    9.0 Gauss's law
    10.0 Work done
    11.0 Electric potential energy
    12.0 Electric Potential
    13.0 Electric dipole

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12.6 Potential Due To Uniformly Charged Spherical Shell

Derivation of expression for potential due to uniformly charge hollow sphere of radius $R$ and Total charge $Q,$ at a point which is at a distance $r$ from a center for the following situation

(i) $r$ $>$ $R$ (ii) $r$ $<$ $R$

Assume a ring of width $Rd\theta $ at angle $\theta$ from $X$ axis (as shown in figure).

Potential due to the ring at the point $P$ will be $$dV = \frac{{K\left( {dq} \right)}}{{\sqrt {{{\left( {r - R\;\cos \theta } \right)}^2} + {{\left( {R\;\sin \theta } \right)}^2}} }}$$

Where $dq = 2\pi R\;\sin \theta \left( {Rd\theta } \right)\sigma $

Where $Q, = 4\pi {R^2}\sigma $

then, net potential $$V = \int {dV = \frac{{KQ}}{2}} \quad \int\limits_0^\pi {\frac{{\sin \theta .\;d\theta }}{{\sqrt {{{\left( {r - R\cos \theta } \right)}^2} + {{\left( {R\sin \theta } \right)}^2}} }}} $$ Solving this eq. we find $$V = \frac{{KQ}}{r}\left( {{\text{for}}\;{\text{r}}\;{\text{ > }}\;{\text{R}}} \right)$$ & $$V = \frac{{KQ}}{R}\left( {{\text{for}}\;{\text{r}}\; < \;{\text{R}}} \right)$$ As the formula of $E$ is easy, we use $V = - \int\limits_{r \to \infty }^{r = r} {\overrightarrow E } .\;d\overrightarrow r $

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(i) At outside point $\left( {r \geqslant R} \right):$ $${V_{out}} = - \int\limits_{r \to \infty }^{r = r} {\left( {\frac{{KQ}}{{{r^2}}}} \right)} \;dr$$ $$ \Rightarrow \;\quad {V_{out}} = \frac{{KQ}}{r} = \frac{{KQ}}{{\left( {{\text{Distance}}\;{\text{from}}\;{\text{centre}}} \right)}}$$ For outside point, the hollow sphere acts like a point charge.

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(ii) Potential at the center of the sphere $(r=0)$

As all the charges are at a distance $R$ from the center,

So, $${V_{center}} = \frac{{KQ}}{R} = \frac{{KQ}}{{\left( {Radius\;of\;the\;sphere} \right)}}$$

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(iii) Potential at inside side point $(r$ $<$ $R):$

Suppose we want to find potential at point $P,$ inside the sphere.

$\therefore $ Potential difference between point $P$ and $O:$ $${V_p} - {V_o} = - \int\limits_O^P {\overrightarrow {{E_{in}}} } .d\overrightarrow r $$ Where, $${E_{in}} = 0$$ So, $${V_p} - {V_o} = 0$$ $$ \Rightarrow \quad {V_p} = {V_o} = \frac{{KQ}}{R}$$ $$ \Rightarrow \quad {V_{in}} = \frac{{KQ}}{R} = \frac{{KQ}}{{\left( {Radius\;of\;the\;sphere} \right)}}$$