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A positive charge q is placed in front o...

A positive charge `q` is placed in front of a conducting solid cube at a distance `d` from its centre. Find the elcetric field at the centre of the cube due to the charges appearing on its surgace.

Text Solution

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Here `E_(i)` = electric field due to induced charges and `E_(q)` = electric field due to charge q
We know that net electric field in a conducting cavity is equal to zero.
Means `vec(E) = 0` at the centre of the cube
`rArr vec(E)_(i) + vec(E)_(q) = 0`
`vec(E)_(i) - vec(E)_(q)`
`vec(E)_(i) = -(kq)/(d^(2))vec(PO)`
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