Two unit negative charges are placed on a straight line. A positive charge $q$ is placed exactly at the…
Two unit negative charges are placed on a straight line. A positive charge $q$ is placed exactly at the midpoint between these unit charges. If the system of these three charges is in equilibrium, the value of $q$ (in $C$) is:
\(1.0\)
\(0.75\)
\(0.5\)
\(0.25\)
Solution
Now total force acting on $(-1)$ at $A$ is
$\begin{aligned}
F_{\text{net}} &= \frac{k(-1)(-1)}{r^{2}} + \frac{k(-1)(q)}{\left(\frac{r}{2}\right)^{2}} = 0 \\
\frac{1}{r^{2}} &= \frac{4q}{r^{2}} \\
q &= \frac{1}{4} = 0.25 C
\end{aligned}$
\(\therefore\) Value of \(q\) for these three charges to be in equilibrium \(=0.25 \mathrm{C}\)
