Question

Two point electric charges of values $q$ and $2q$ are kept at a distance $d$ apart from each other in air. A third charge $Q$ is to be kept along the same line in such a way that the net force acting on $q$ and $2q$ is zero. Calculate the position of charge $Q$ in terms of $d$.

• A. $\frac{d}{\sqrt{2}-1}$
• B. $\frac{d}{\sqrt{2}+1}$
• C. $\frac{d}{\sqrt{3}+1}$
• D. $\frac{d}{\sqrt{3}-1}$

Answer 1

Answer: (B)

Suppose that the charge $q,2q$ and $Q$ are placed as shown in the figure.

It follows that the net force on charge $q$ and $2q$ can be zero, only if the charge $Q$ is of opposite sign to those of charges $q$ and $2q$. Therefore, if charges $q$ and $2q$ are positive, then charge $Q$ must be negative in nature. Let the distance of charge $Q$ (negative) from $q$ be equal to $x$.
For force on charge $q$ to be zero, $F_{AB}=F_{AC}$
or $\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q(2q)}{d^2}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{qQ}{x^2}$
or $\frac{Q}{q}=\frac{2x^2}{d^2}$ (i)
For force on charge $2q$ to be zero, $F_{BA}=F_{BC}$
or $\frac{1}{4\pi {\epsilon }_0}\cdot \frac{q(2q)}{d^2}=\frac{1}{4\pi {\epsilon }_0}\cdot \frac{(2q)Q}{(d-x{)}^2}$
or $\frac{Q}{q}=\frac{(d-x{)}^2}{d^2}$ (ii)
From the equations (i) and (ii), we get
$x=\frac{d}{\sqrt{2}+1}$ or $-\frac{d}{\sqrt{2}-1}$