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  $E^{\mathrm{\prime }}=\frac{Q}{x^2}=\frac{E}{3}$ (along AB directed away from positive charge