Question
A positive charge q is placed in front of a conducting solid cube at a distance d from its centre. Find the electric field at the centre of the cube due to the charges appearing on its surface.
We know,
Electric field at a point due to a given charge
$'\text{E}'=\frac{\text{Kq}}{\text{r}^2}$ Where q = charge, r = Distance between the point and the charge
So, $'\text{E}'=\frac{1}{4\pi\in_0}\times\frac{\text{q}}{\text{d}^2}$ $[\therefore\text{r}=\text{‘d’}\text{here}]$
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