First slide

Gauss's Law

Question

It has been experimentally observed that the electric field in a large region of earths atinosphere is directed vertically down. At an altitude of 300 m, the electric field is 60 vm^-1 . At an altitude of 200 m, the field is 100 vm^-1. Find the net amount of charge contained in the cube of 100 m edge, located between 200 m and 300 m altitude is

Moderate

A

$1.5 \times 10^{5} ε_{0} C$
B

$2 \times 10^{5} ε_{0} C$
C

$3 \times 10^{5} ε_{0} C$
D

$4 \times 10^{5} ε_{0} C$

Solution

According to Gauss' theorem, electric flux: $ϕ_{E} = \frac{q}{ε_{0}} = \int \vec{E} \cdot d \vec{S}$
The surface integral in the above equations contains six terms - the surface integral over the bottom surface, the surface integral over the top surface and surface integral over the four vertical faces . For the bottom surface, both the vectors $\vec{E} and \vec{dS}$ are in the same direction. For the top surface, they act in opposite directions while for the vertical aces, they are perpendicular to each other.
$\begin{array}{rcl} Hence, ϕ_{E} & = & \int_{bottom} {\vec{E}}_{1} \cdot d \vec{S} + \int_{top} {\vec{E}}_{2} \cdot d \vec{S} + 4 \int_{faces} \vec{E} . d \vec{S} \\ = & \int_{bottom} E_{1} \cdot dScos 0 ° + \int_{top} E_{2} \cdot dScos 180 ° + 4 \int_{faces} EdScos 90 ° \\ = & \int_{bottom} E_{1} \cdot dS - \int_{top} E_{2} \cdot dS = E_{1} S - E_{2} S = (E_{1} - E_{2}) S \\ Also, ϕ_{E} & = & \frac{q_{enclosed}}{ε_{0}} \\ q & = & ε_{0} (E_{1} - E_{2}) S = ε_{0} (100 {Vm}^{- 1} - 60 {Vm}^{- 1}) (100 m)^{2} \\ = & 4 \times 10^{5} ε_{0} C \end{array}$

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