Question

A hollow cylinder has a charge q coulomb within it. It ϕ is the electric flux associated with the curved surface B, the flux linked with the plane surface A will be

• A. $\frac {ϕ}{3}$
• B. $\frac {q}{ε_0} - ϕ$
• C. $\frac {1}{2}(\frac {q}{ε_0} - Φ)$
• D. Zero

Answer 1

Answer: (C)

$\frac {1}{2}(\frac {q}{ε_0} - Φ)$

Answer 2

The flux linked with a closed surface is given by Gauss's Law.
Φ = $\frac {q}{ε₀}$
In the given scenario, there is a charge q within the hollow cylinder. Let's consider the curved surface B and the plane surface A separately.
For the curved surface B, the electric flux associated with it is ϕ.
For the plane surface A, the flux linked with it can be calculated as follows:
Φ(A) = Φ - Φ(B)
Substituting the values, we get:
Φ(A) = $\frac {q}{ε₀}$ - ϕ
Therefore, the correct answer is (C) $\frac {1}{2}(\frac {q}{ε_0} - Φ)$.