9.6 Electric field due to a solid sphere of charge

  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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9.6 Electric field due to a solid sphere of charge

The electric field at any point which is at a distance $r$ from the centre of solid sphere of radius $R$ and charge $q$ which is uniformly distributed throughout the volume of solid sphere is given as follows,

$$\begin{equation} \begin{aligned} {E_{{\text{centre}}}} = 0 \\ {E_{{\text{inside}}}} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{R^3}}}r \\ {E_{{\text{outside}}}} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{r^2}}} \\\end{aligned} \end{equation} $$

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Proof: Electric field inside the solid sphere

Consider a positive charge $q$ is uniformly distributed throughout the volume of a solid sphere of radius $R$.

So, the charge per unit volume is given by, $\rho = \frac{q}{{\frac{4}{3}\pi {R^3}}}$

For finding the electric field at a distance $r\ (<R)$ from the centre, we will draw a gaussian surface in the form of sphere of radius $r$ and concentric with the charge distribution.

At all point, the electric field is radial and is perpendicular to the surface. So, from the Gauss's law, we can write,
$$ES = \frac{{{q_{in}}}}{{{\varepsilon _0}}}$$
where,

$S = 4\pi {r^2}:$ Surface area of the gaussian surface
${q_{in}} = \rho \left( {\frac{4}{3}\pi {r^3}} \right) = \left( {\frac{q}{{\frac{4}{3}\pi {R^3}}}} \right)\left( {\frac{4}{3}\pi {r^3}} \right) = \frac{{q{r^3}}}{{{R^3}}}:$ Charge enclosed within the gaussian surface.
$$E\left( {4\pi {r^2}} \right) = \frac{{\left( {\frac{{q{r^3}}}{{{R^3}}}} \right)}}{{{\varepsilon _0}}}$$$$E = \frac{1}{{4\pi {\varepsilon _0}}}.\frac{q}{{{R^3}}}r$$
Hence, the electric field inside the solid sphere is directly proportional to the distance from the centre.
$$E \propto r$$

Electric field at the centre of the solid sphere,
$$E = 0\quad \left( {{\text{As, }}r = 0} \right)$$
Electric field at the surface of the solid sphere,
$$E = \frac{1}{{4\pi {\varepsilon _0}}}\frac{q}{{{R^2}}}\quad \left( {{\text{As, }}r = R} \right)$$

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Proof: Electric field outside the solid sphere

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