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A Gaussian surface in the cylinder of cr...

A Gaussian surface in the cylinder of cross-section `pia^(2)` and length L is immersed in a uniform electric field E with the cylinder axis parallel to the field. The flux `phi` of the electric field through the closed surface is

A

(a) `2 pia^(2)E`

B

(b) `pia^(2)EL`

C

(c) `pia^(2)(2+L) E`

D

(d) zero

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The correct Answer is:
To find the electric flux \(\Phi\) through a closed cylindrical Gaussian surface immersed in a uniform electric field \(E\), we can follow these steps: ### Step 1: Understand the Geometry The cylindrical Gaussian surface has a cross-sectional area \(A = \pi a^2\) and length \(L\). The electric field \(E\) is uniform and directed parallel to the axis of the cylinder. ### Step 2: Identify the Surfaces The closed surface of the cylinder consists of: - Two circular ends (cross-sections) - A curved lateral surface ### Step 3: Analyze the Flux Through Each Surface 1. **Curved Surface**: The electric field is parallel to the axis of the cylinder, while the area vectors of the curved surface are perpendicular to the electric field. Therefore, the flux through the curved surface is zero: \[ \Phi_{\text{curved}} = 0 \] 2. **End Surface 1**: For one circular end (let's call it end 1), the area vector is directed opposite to the electric field. The angle between the electric field vector and the area vector is \(180^\circ\): \[ \Phi_1 = -E \cdot A = -E \cdot (\pi a^2) \] 3. **End Surface 2**: For the other circular end (end 2), the area vector is in the same direction as the electric field. The angle between the electric field vector and the area vector is \(0^\circ\): \[ \Phi_2 = E \cdot A = E \cdot (\pi a^2) \] ### Step 4: Calculate the Total Flux The total electric flux through the closed surface is the sum of the fluxes through both ends and the curved surface: \[ \Phi = \Phi_1 + \Phi_2 + \Phi_{\text{curved}} = (-E \cdot \pi a^2) + (E \cdot \pi a^2) + 0 \] \[ \Phi = -E \cdot \pi a^2 + E \cdot \pi a^2 = 0 \] ### Conclusion The total electric flux \(\Phi\) through the closed cylindrical surface is: \[ \Phi = 0 \]
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