First slide

Gauss's Law

Question

In the arrangement shown, electric flux through surface 1 is $ϕ$ . Then what will be the electric flux through the closed surface 2 ?

Moderate

A

$(ϕ + 10^{4}) v - m$
B

$(ϕ + 10^{3}) v - m$
C

$(ϕ - 10^{4}) v - m$
D

$(ϕ + 10^{6}) v - m$

Solution

By Guass's law
$Q_{1} + Q_{2} = ε_{0} x ϕ - - - (1) and Q_{1} + Q_{2} + 8.85 x 10^{- 6} = ε_{0} x ϕ^{1} - - - (2) ∴ (2) - (1) \Rightarrow ε_{0} x ϕ^{1} - ε_{0} x ϕ = 8.85 x 10^{- 6} \Rightarrow ϕ = ϕ + \frac{8.85 x 10^{- 6}}{8.85 x 10^{- 12}} = (ϕ + 10^{6}) v - m$

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