Question

Obtain an expression for electric field intensity due to a uniformly charged sphere.​

Answer 1

Electrostatic Field Intensity due to a spherical shell.

• Let's assume that a spherical shell has charge q on its surface and radius R

1. Now, for a point r greater than the radius of sphere (i.e. r > R)

• Applying GAUSS' LAW:

$\displaystyle \: \oint \: E \times ds = \dfrac{q_{enclosed} }{ \epsilon_{0}}$

$\implies \displaystyle \:E \oint ds = \dfrac{q }{ \epsilon_{0}}$

$\implies \displaystyle \:E (4\pi {r}^{2} )= \dfrac{q }{ \epsilon_{0}}$

$\implies \displaystyle \:E = \dfrac{q }{ 4 \epsilon_{0}\pi {r}^{2} }$

2. For a point on surface of sphere (r = R):

$\displaystyle \: \oint \: E \times ds = \dfrac{q_{enclosed} }{ \epsilon_{0}}$

$\implies \displaystyle \:E \oint ds = \dfrac{q }{ \epsilon_{0}}$

$\implies \displaystyle \:E (4\pi {R}^{2} )= \dfrac{q }{ \epsilon_{0}}$

$\implies \displaystyle \:E = \dfrac{q }{ 4 \epsilon_{0}\pi {R}^{2} }$

3. For a point inside sphere (when r < R):

• In this case, no charge is enclosed as all charge is in surface.

$\displaystyle \: \oint \: E \times ds = \dfrac{q_{enclosed} }{ \epsilon_{0}}$

$\implies \displaystyle \: \oint \: E \times ds = \dfrac{0}{ \epsilon_{0}}$

$\implies \displaystyle E = 0$

Hope It Helps.