A point charge of magnitude q is at the center of a cube with sides of length L

Question

Question:

A point charge of magnitude q is at the center of a cube with sides of length L

(A) What is the electric flux Φ through each of the six faces of the cube?

(B) What would be the flux Φ1 through a face of the cube if its sides were of length L1?

Answer:

Gauss’ Law:

The complete electric transition through any shut surface can be effectively discover by the utilization of Gauss’ law. As per Gauss’ law, the complete electric motion through any shut surface is communicated as:
ϕ =enclosed chargeϵoϕ =enclosed chargeϵo
Motion out of a Cube
Section A = Φ = q/(6ε0)
Part B = Φ = q/(6ε0)

Arrangements Below:

A point charge of extent q is at the focal point of a 3D square with sides of length L.

Section A

What is the electric motion Φ through every one of the six essences of the solid shape?
Use ε0 for the permittivity of free space.
As per Gauss’ Law, an inside charge is equivalent to the motion times the permittivity of free space. At the end of the day, we have the recipe:
ε0φ = q
Where ε0 is the permittivity of free space, Φ is the motion, and q is within charge. Since the issue gets some information about the transition at every one of the 6 essences of the solid shape, we simply separate this by 6 to find the solution. We can do this in light of the fact that each face of a solid shape is equivalent. In the event that we were discussing a rectangular box or some other shape that wasn’t balanced, we would need to mull over that, yet not for this issue!
ε0φ = q
Φ = q/ε0
Φface = q/(6ε0)
We could likewise fathom by realizing that the absolute transition is equivalent to the electric field at a specific point inside a region, times the all out region. So since each face of the solid shape is 1/sixth it’s surface territory, the complete transition on a given face would be given by:
Φ = E * 1/6A
Yet, at that point we would need to relate E to q, which is past the extent of this issue (however you can discover it arithmetically in the event that you need).
q/(6ε0)

Part B
What might be the motion Phi1 through a face of the solid shape if its sides were of length L1?
Use ε0 for the permittivity of free space.
The appropriate response is equivalent to for Part A. The length of the sides doesn’t make a difference on the grounds that a 3D shape has equivalent countenances and each face is constantly 1/sixth of the absolute surface region.
q/(6ε0)

 

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Sam Professor 2 years 1 Answer 1445 views 0

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