Question

Two point electric charges of values $q$ and $2 q$ 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 $2 q$ 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 $q, 2 q$ and $Q$ are placed as shown in the figure.

It follows that the net force on charge $q$ and $2 q$ can be zero, only if the charge $Q$ is of opposite sign to those of charges $q$ and $2 q$. Therefore, if charges $q$ and $2 q$ 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, $F_{A B}=F_{A C}$
or $\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q(2 q)}{d^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q Q}{x^{2}}$
or $\frac{Q}{q}=\frac{2 x^{2}}{d^{2}}$ (i)
For force on charge $2 q$ to be zero, $F_{B A}=F_{B C}$
or $\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q(2 q)}{d^{2}}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{(2 q) Q}{(d-x)^{2}}$
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 $-\frac{d}{\sqrt{2}-1}$