Q. Charges $4\, Q,\, q$ and $Q$ are placed along x-axis at position $ x=o, x=l/2 $ and $ x=l, $ respectively. Find the value of $q$, so that force on charge $Q$ is zero.
Delhi UMET/DPMTDelhi UMET/DPMT 2005Electric Charges and Fields
Solution:
From Coulombs law, the force acting between two charges $\left(q_{1}, q_{2}\right)$
separated at a distance $r$ is given by
$F=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r^{2}}$
Total force acting on charge $Q$ is
$F= \frac{1}{4 \pi \varepsilon_{0}} \frac{q Q}{(l / 2)^{2}}+\frac{1}{4 \pi \varepsilon_{0}} \frac{4 Q \cdot Q}{(l)^{2}}$
According to question, $F =0$
$\therefore \frac{1}{4 \pi \varepsilon_{0}} \frac{q Q}{(l / 2)^{2}}+\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{4 Q^{2}}{(l)^{2}}=0$
$\Rightarrow \frac{4 q}{l^{2}}+\frac{4 Q}{l^{2}}=0$
