First slide

Gauss's Law

Question

In the diagram shown, electric flux over the surface of Gaussian surface (1) is $8 \times 10^{4} V - m$ and the flux over the Gaussian surface (2) is $12 \times 10^{4} V - m$ . Then the charge $Q_{1}$ is equal to

Moderate

A

0.885 $μ C$
B

1.475 $μ C$
C

-0.56 $μ C$
D

0.234 $μ C$

Solution

By Gauss's law, $Q_{1} - Q_{2} = \in_{0} ϕ_{1} and Q_{1} + Q_{2} = \in_{0} ϕ_{2}$
$∴ Q_{1} = \frac{1}{2} \in_{0} (ϕ_{1} + ϕ_{2}) = \frac{1}{2} \times 8.85 \times 10^{- 12} \times 20 \times 10^{4} C = 0.885 μ C$

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