Question

A charge q is placed at the centre of the open end of a cylindrical vessel. The flux of the electric-field through the surface of the vessel is

• A. zero
• B. \(\frac{q}{\epsilon_0}\)
• C. \(\frac{q}{2\epsilon_0}\)
• D. \(\frac{2q}{\epsilon_0}\)

Hint:

Whenever you encounter problems with open surfaces and need to find the electric flux, try to create a symmetrical closed surface by adding an imaginary part. Apply Gauss's Law to the closed surface and then use symmetry to determine the flux through the original open part.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This problem is based on Gauss's Law of electrostatics. Gauss's Law states that the total electric flux (\(\Phi_E\)) through any closed surface (known as a Gaussian surface) is equal to \(\frac{1}{\epsilon_0}\) times the net electric charge (\(q_{enc}\)) enclosed by the surface.
The surface of the cylindrical vessel given in the problem is an open surface, not a closed one. Therefore, we cannot directly apply Gauss's law to it.

Step 2: Key Formula or Approach:
The key formula is Gauss's Law: \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0} \] To use this law, we must construct a hypothetical closed surface (a Gaussian surface) for which the flux can be calculated, and then use symmetry to find the flux through the required open surface.

Step 3: Detailed Explanation:
Let's consider the open cylindrical vessel. The charge \(q\) is placed at the center of its open end.
To create a closed surface, imagine an identical cylindrical vessel placed symmetrically on top of the first one, closing the open end.
This combination forms a closed cylinder, and the charge \(q\) is now located at the center of this closed cylinder.
According to Gauss's Law, the total electric flux through this entire closed cylindrical surface is: \[ \Phi_{total} = \frac{q}{\epsilon_0} \] The closed surface consists of the original vessel and the identical imaginary vessel. Due to the symmetry of the charge's position, the electric field lines will pass equally through both halves of the closed cylinder.
Therefore, the flux through the original open cylindrical vessel will be exactly half of the total flux.
\[ \Phi_{vessel} = \frac{\Phi_{total}}{2} = \frac{1}{2} \left( \frac{q}{\epsilon_0} \right) = \frac{q}{2\epsilon_0} \]

Step 4: Final Answer:
The flux of the electric-field through the surface of the vessel is \(\frac{q}{2\epsilon_0}\).