Name: Electric Field of a Charged Plane Sheet
Uploaded: 2019-07-19
Duration: 6 min 58 s
Description: Another distribution of charge. This time on an infinite plane sheet. Let’s see once again how Gauss’s law comes to the rescue in electric field calculations.

Question

A hemispherical body is placed in a uniform electric field $E$ . What is the flux linked with the curved surface if the field is $(i)$ parallel to the base of the body [Figure (a)]; $(ii)$ perpendicular to base of the body [Figure (b)] and $(iii)$ perpendicular to the curved surface at every points as in Figure (c).

Accepted Answer

The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field. In integral form it is given by,

$(i)$ , where $\vec{A}$ is the area of the closed surface and $\vec{E}$ is the electric field.

Electric flux is defined as the number of electric field lines passing through a closed surface.

Let the $ϕ_{1}$ is the flux passing through the curved surface and $ϕ_{2}$ is the flux passing through the plane surface.

Case (a):

As we can see that total electric field lines entering the curved surface is equal to the total electric field lines leaving the surface, therefore, net electric flux will be zero.

$ϕ_{1} = 0 \dots (1)$ .

When the field is parallel to the base surface,

$ϕ_{2} = \vec{E} \cdot \vec{A} = E \times π R^{2} \cos 90^{\circ} = 0$

Using equation $(1)$ , $ϕ_{net} = ϕ_{1} + ϕ_{2} = 0$ .

Case (b):

As we can see that total electric field lines entering the plane surface is equal to the total electric field lines leaving the curved surface, therefore, net electric flux will be zero. $\Rightarrow ϕ_{1} = 0$ .

When the field is perpendicular to the plane surface,

$ϕ_{2} = \vec{E} \cdot \vec{A} = E \times π R^{2} \cos 180^{\circ} \Rightarrow ϕ_{2} = - E π R^{2}$

$ϕ_{net} = ϕ_{1} + ϕ_{2} = E π R^{2}$ .

Case (c):

When the electric field lines are perpendicular to the curved surface,

$ϕ_{1} = \vec{E} \cdot \vec{A} = E \times 2 π R^{2} \cos 0^{\circ}$

$ϕ_{1} = 2 E π R^{2}$

Important Questions on Electric Flux and Gauss's Law

Learn and Prepare for any exam you want !