State Gauss law in electrostatics. Using the law derive an expression for electric field due to a uniformly charged thin spherical shell at a point outside the shell.

Gauss’s law relates the flux through a closed surface (a surface that encloses some volume) with the electric charges present inside the surface.

According to Gauss's law, the total electric flux (Φ) through any closed surface that surrounds a charge (q) in free space is equal to the charge divided by the absolute permittivity (∈_o). 

Φ = \(\frac{q}{ \in _0}\)

Since, electric flux, \(\phi = \oint _s \overrightarrow{E} . \overrightarrow{ds}\)

Therefore, Gauss’s law can be written as

\(\phi = \oint _s \overrightarrow{E} . \overrightarrow{ds} = \frac{q}{ \in _0}\)

Expression for electric field intensity due to a uniformly charged thin spherical shell at a point outside the shell

Consider a thin spherical shell of radius R having a charge Q uniformly distributed on its surface.

We will find electric field intensity at distance r from the center of the shell, such that r > R. We enclose the shell in a gaussian sphere of radius r.

According to Gauss’s theorem, the net electric flux through the gaussian surface is given by

\(\phi = \oint _s \overrightarrow{E} . \overrightarrow{ds} = \frac{Q}{ \in _0}\Rightarrow \oint _s Eds cos \theta =\frac{Q}{ \in _0} \)

Direction of the electric field is always perpendicular to the surface. So, the angle between E and ds is 0.

⇒ \(\oint _s\) Eds cos0 = \(\frac{Q}{ \in _0} \)

⇒ \(\oint _s\)Eds = \(\frac{Q}{ \in _0} \)

Since the electric field is constant at every point of the gaussian surface, therefore we can write

E\(\oint _s\)ds = \(\frac{Q}{ \in _0} \)

But, \(\oint _s\)ds  = 4πr² is the surface area of sphere

⇒ E x 4πr² = \(\frac{Q}{ \in _0} \)

⇒ E = \(\frac{Q}{4 \pi \in_o r^2}\)

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