A positive charge q is placed in front of a solid conducting cube at a distance d from its centre. Calculate the electric field at the centre of the cube due to the charges appearing on its surface.

chetan jangir
101 Points
5 years ago
In this problem there is a very beautiful application of fact that the electric field inside a conductor is zero.As now we can conclude that the net electric field inside the conductor is zero due to all the charges that are in the given problem.We can calculate net electric field at center and put it equal to zero,So net field is due to charge q and charge appearing on surface of conductor . $E_{q} +E_{induced charge}=0$.We can find E due to q that is simply kq/$d^{2}$ .So electric field due to induced charge will have magnitude kq/$d^{2}$ and diection opposite to field due to q
Thanks..Hope you Understand
Rajdeep
231 Points
5 years ago
HELLO!

We know, that electric field at the centre of a conductor is zero.
So, when we place a point charge at a distance d from the centre, then the electric field at the centre will be due to the electric field due to the point charge. (Read this sentence twice if you don’t understand).

So,
$E = \frac{Kq}{d^{2}} = \frac{q}{4\pi \epsilon_{o}d^{2}}$
chetan jangir
101 Points
5 years ago
Hey Rajdeep pls read question carefully we are not asked field due to charge.We were asked field due to charge induced on cube.I think you are going through  answer that's why you are explaning the correct final answer through a wrong way.That is not the Physics.