First slide

Gauss's Law

Question

A hollow cylinder has a charge q coulomb within it. If $ϕ$ is the electric flux in unit of volt-meter associated with the curved surface B, the flux linked with the plane surface A in unit of volt-metre will be

Moderate

A

$\frac{q}{ε_{0}} - ϕ$
B

$\frac{q}{2 ε_{0}}$
C

$\frac{ϕ}{3}$
D

$\frac{1}{2} (\frac{q}{ε_{0}} - ϕ)$

Solution

Gauss's law states that the net electric flux through any closed surface is equal to the net charge inside the surface divided by ε₀·
$i.e., ϕ_{total} = \frac{q}{ε_{0}}$
Let electric flux linked with surfaces A, B and C are $ϕ_{A}, ϕ_{B} and ϕ_{C}$ , respectively. That is
$ϕ_{total} = ϕ_{A} + ϕ_{B} + ϕ_{C} Since ϕ_{C} = ϕ_{A} ∴ 2 ϕ_{A} + ϕ_{B} = ϕ_{total} = \frac{q}{ε_{0}} or ϕ_{A} = \frac{1}{2} (\frac{q}{ε_{0}} - ϕ_{B}) But ϕ_{B} = ϕ (given) Hence, ϕ_{A} = \frac{1}{2} (\frac{q}{ε_{0}} - ϕ)$

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