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

The Correct Option is C

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) \(\frac {1}{2}(\frac {q}{ε_0} - Φ)\).