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Question: consider a charge placed at a distance a/2 from the centre of a cube of side length a. Find flux thr...

consider a charge placed at a distance a/2 from the centre of a cube of side length a. Find flux through each face of cube

Answer

Flux through the face where the charge is placed = 0 Flux through each of the other five faces = q/(10ε₀)

Explanation

Solution

  1. Identify the charge location: The charge 'q' is placed at a distance a/2a/2 from the center of the cube of side length 'a'. This means the charge is located at the center of one of the faces of the cube.

  2. Total flux through the cube: Since the charge is on the surface of the cube, only half of the total flux originating from the charge passes through the cube. Φcube=12(qϵ0)=q2ϵ0\Phi_{cube} = \frac{1}{2} \left( \frac{q}{\epsilon_0} \right) = \frac{q}{2\epsilon_0}

  3. Flux through the face containing the charge: If a point charge is placed exactly on a surface, the electric flux through that surface is considered to be zero. Φface_with_charge=0\Phi_{face\_with\_charge} = 0

  4. Flux through the remaining five faces: The entire flux Φcube=q/(2ϵ0)\Phi_{cube} = q/(2\epsilon_0) must pass through the other five faces. In the absence of specific information requiring complex integration or detailed solid angle calculations, it is often assumed that this flux is equally distributed among these five faces due to symmetry. Flux through each of the other five faces = Φcube5=q/(2ϵ0)5=q10ϵ0\frac{\Phi_{cube}}{5} = \frac{q/(2\epsilon_0)}{5} = \frac{q}{10\epsilon_0}

Final Answer:

  • Flux through the face where the charge is placed = 00
  • Flux through each of the other five faces = q10ϵ0\frac{q}{10\epsilon_0}

Explanation of the solution:

  1. The charge is at the center of one face.
  2. Total flux from charge is q/ϵ0q/\epsilon_0. Half of this flux enters the cube: Φcube=q/(2ϵ0)\Phi_{cube} = q/(2\epsilon_0).
  3. Flux through the face containing the charge is zero because field lines are tangential to the surface over its area.
  4. The remaining flux q/(2ϵ0)q/(2\epsilon_0) passes through the other five faces.
  5. Assuming equal distribution among these five faces due to symmetry, flux through each of these faces is 15×q2ϵ0=q10ϵ0\frac{1}{5} \times \frac{q}{2\epsilon_0} = \frac{q}{10\epsilon_0}.