Question

electric field intensity depends on the nature of Gaussian

surface? Explain.​

Answer 1

Gaussian surface is an area vector

Answer 2

To explain:

Electric field intensity depends on the nature of Gaussian surface .

Explaination:

Gauss' Law states that the net flux through to a surface is equal to the enclosed charge divided by the permittivity of free space.

$\boxed{ \sf{ \int\vec{E} \: . \: \vec{ds} = \frac{q}{ \epsilon_{0}} }}$

Considering the angle between field intensity factor and the area vector to be $\theta$.

$= > \displaystyle \sf{ \int E \times ds \times \cos( \theta) = \frac{q}{ \epsilon_{0}} }$

$= > \displaystyle \sf{ E\int ds = \frac{q}{ \epsilon_{0} \cos( \theta) } }$

$= > \displaystyle \sf{ E \times s = \frac{q}{ \epsilon_{0} \cos( \theta) } }$

$= > \displaystyle \sf{ E = \frac{q}{ \epsilon_{0} \cos( \theta) s} }$

Hence , field intensity is dependent on the type of surface area enclosing the charge.

• During calculations we usually take standard geometrical figures as Gaussian surfaces.

• For example : for line charge we consider a cylindrical Gaussian surface ; for point charge , a spherical Gaussian surface is considered.