Two-point charges Q and $– 3Q$ are placed at some distance apart. If the electric field at the location of Q  is E then at the locality of $– 3Q$ , it is.

The field produced by charge – 3Q at A, this is E as mentioned in the

 Example.

$\therefore$$E=\frac{3Q}{x^2}$
  (alongAB directed towards negative charge)

Now field at location of – 3Qi.e. field at B due to charge Q will be  $E\text{'}=\frac{Q}{x^2}=\frac{E}{3}$ (along AB directed away from positive charge