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

At each point on the surface of the cube shown in the figure, the electric field is parallel to the $z$ axis. The length of each edge of the cube is $4.0 m$ . On the top face of the cube the field of $\vec{E} = - 34 \hat{k} N C^{- 1}$ and on the bottom face it is $\vec{E} = + 20 \hat{k} N C^{- 1}$ . Determine the net charge contained within the cube.

Accepted Answer

Given, direction of electric is along z-direction.

Electric field on the top surface, ${\vec{E}}_{t o p} = - 34 \hat{k} N C^{- 1}$

Electric field on the bottom surface, ${\vec{E}}_{b o t} = + 20 \hat{k} N C^{- 1}$

Edge of the cube, $a = 4.0 m$

Gauss's law

$ϕ = \underset{s}{∯} {\vec{E}}_{1} \cdot d \vec{S} = \frac{q_{e n c}}{ε_{0}}$

So, Net flux linked with the cube is given by,

$ϕ = \iint_{t o p} \vec{E} \cdot d \vec{S} + \iint_{b o t t o m} \vec{E} \cdot d \vec{S} + \iint_{o t h e r} \vec{E} \cdot d \vec{S}$

The cube has $6$ surfaces and except top and bottom surfaces, for other surfaces flux linked become zero. Because electric field is along z-direction and normal vector to the surface has no component along z-direction.

So,

$ϕ = \iint_{t o p} \vec{E} \cdot d \vec{S} + \iint_{b o t t o m} \vec{E} \cdot d \vec{S}$

$\iint_{t o p} \vec{E} \cdot d \vec{S} = {\vec{E}}_{t o p} \cdot {\vec{A}}_{t o p} + {\vec{E}}_{b o t t o m} \cdot {\vec{A}}_{b o t t o m} = (- 34 \hat{k}) \cdot (a^{2} \hat{k}) + (- 20 \hat{k}) \cdot (a^{2} \hat{k}) = - 864 N m^{2} C^{- 1}$

(taking inward flux as negative)
Net charge contained within the cube

$q_{enc} = ε_{0} ϕ = (8.85 \times 10^{- 12} C^{2} N^{- 1} m^{- 2}) \times (- 864 N m^{2} C^{- 1}) = 7.65 \times {10^{- 9} N m^{2} C}^{- 1}$

Important Questions on Gauss' Law

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