91Ó°ÊÓ

We are given two charges +Q at the origin and +4Q at distance d from +Q.

Each charge has an electric field, and as both of them have a positive charge, the electric field from each charge is away from the charge, so the place of the charge q will be between the two charges +Q and +4Q because the two charges will be in opposite direction and the net electric field will be zero at a certain distance between them, let us consider the distance between the charge q and +Q is x so the distance between the charge q and +4Q will equal (d-x) as shown below.

Calculate x to determine the place of charge q, where the two electric fields at this point have the same magnitude.

Let E1 for charge+Q and E2 charge+4Q.

$\begin{array}{r}{E}_1=E_2\\ k\frac{\left|Q\right|}{x^2}=k\frac{\left|4Q\right|}{(\mathrm{d}−\mathrm{x}{)}^2}\\ \frac{1}{x^2}=\frac{4}{(\mathrm{d}−\mathrm{x}{)}^2}\\ \frac{d−x}{x}=2\\ \frac{d}{x}=3\\ x=\frac{d}{3}\end{array}$

Hence, the charge q is placed at distance d/3 where the electric field is zero.

As we know that charges q and +4Q also exerted a force on the charge +Q, where this force is at equilibrium as the electric field is zeroand it is given by

$\begin{array}{r}{F}_q+F_2=0\\ F_q=−F_2\\ k\frac{qQ}{x^2}=−k\frac{4QQ}{d^2}\\ q=−\frac{4Q{x}^2}{d^2}\\ =−\frac{4Q{\left(\frac{d}{3}\right)}^2}{d^2}\\ =−\frac{4Q}{9}\end{array}$

Hence the magnitude of q is$\frac{4Q}{9}$ and its sign is negative.