Rankers Physics

Electric Flux: Practice Problem & Solution

Figure shows four charges q1, q2, q3 and q4 fixed in space. Then the total flux of electric field through a closed surface S, due to all charges q1, q2, q3, and q4 is : Electric Flux diagram: Figure shows four charges q1, q2, q3 and
not equal to the total flux through S due to charges q3 and q4.
equal to the total flux through S due to charges q3 and q4.
zero if q1 + q2 = q3 + q4
twice the total flux through S due to charges q3 and q4 if q1 + q2 = q3 + q4

Solution Explained:

To solve this problem, we apply the core principles of Electric Flux. Understanding the underlying formula is key to arriving at the correct answer below:

Gauss's law states that the total electric flux passing through a closed surface is equal to the charge enclosed within the surface divided by the permittivity of the medium.

In short: The net electric field around a closed surface depends on the charge inside it.

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. A charge \(Q\) \(mu\text{C}\) is placed at the centre of a cube. The flux coming out from any one of its faces will be (in SI unit)
2. The electric field components in the given figure are \(  E_{x}=\alpha x^{1/2}, E_{y}=E_{z}=0 \) in which \( \alpha=800NC^{-1}m^{-1/2}.\) The charge within the cube is if net flux through the cube is 1.05 N m²C–¹ (assume a = 0.1 m)  
3. The electric flux Φ through a hemisphere surface of radius R, placed in a uniform electric field of intensity E parallel to the axis of its circular plane is:
4. The length of each side of a cubical closed surface is l. If charge q is situated on one of the vertices of the cube, then find the flux passing through shaded face of the cube.
5. A charge Q is placed at the mouth of a conical flask. The flux of the electric field through the flask is