A charge Q = 10 nC is placed at the origin r = (0, 0, 0), at the centre of a cub

Question

A charge Q = 10 nC is placed at the origin r = (0, 0, 0), at the centre of a cube of side-length a = 1 cm whose faces are perpendicular to the x, y, and z-axes as shown below left. What is the total electric flux Phi_E through the box, and what is the average of the perpendicular component of the electric field, E on the right-hand face (the one centered on (0,a/2,0))? A second charge, - Q, is now added at r = (0, a, 0), as shown on the right. Now what is the total flux through the box, and the average of the perpendicular component, E on the face centered on (0, a/2,0)?

Explanation / Answer

(a)
According to Gauss Law, The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.

So,
E = q/eo
Total Electric Flux through the box, E = (10 * 10^-9)/(8.85*10^-12) Vm
E = 1130 Vm

E = Eperpendicular * A
Eperpendicular = E/A
Eperpendicular = 1130/(1*10^-4) N/C
Eperpendicular = 1130 * 10^4 N/C

(B)
As the Second charge is not Enclosed in the box, Therefore Total Flux through the box remains the same .
E = 1130 Vm

In this Case,
Eperpendicular = E/A
Eperpendicular = 2*Q/(eo*A)
Eperpendicular = 2* (10 * 10^-9)/((8.85*10^-12) * 10^-4)
Eperpendicular = 2260 * 10^4 N/C

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