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 $\mathrm{nE}$. The value of $\mathrm{n}$ is

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

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

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