b
(b) The field produced by charge \(-3Q\) at \(A\), this is \(E\) as mentioned in the Example.
\(\therefore \) \(E = \frac{{3Q}}{{{x^2}}}\) (along \(AB\) directed towards negative charge)
Now field at location of \(-3Q\) i.e. field at \(B\) due to charge \(Q\) will be \(E' = \frac{Q}{{{x^2}}} = \frac{E}{3}\) (along \(AB\) directed away from positive charge)
