Electrostatics Ques 52
- Three point charges $q, 2 q$ and $8 q$ are to be placed on a $9 \mathrm{~cm}$ long straight line. Find the positions where the charges should be placed such that the potential energy of this system is minimum. In this situation, what is the electric field at the position of the charge $q$ due to the other two charges?
$(1987,7 M)$
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Answer:
Correct Answer: 52.(a) Charge $q$ should be at a distance of $3 \mathrm{~cm}$ from $2 q$
(b) Electric field $=0$
Solution:
For potential energy to be minimum the bigger charges should be farthest. Let $x$ be the distance of $q$ from $2 q$. Then potential energy of the system shown in figure would be
$$ U=K \frac{(2 q)(q)}{x}+\frac{(8 q)(q)}{(9-x)}+\frac{(2 q)(8 q)}{9} $$
Here, $K=\frac{1}{4 \pi \varepsilon_{0}}$
For $U$ to be minimum $\frac{2}{x}+\frac{8}{9-x}$ should be minimum.
$$ \begin{array}{rc} & \frac{d}{d x} \frac{2}{x}+\frac{8}{9-x}=0 \\ \therefore & \frac{-2}{x^{2}}+\frac{8}{(9-x)^{2}}=0 \\ \therefore & \frac{x}{9-x}=\frac{1}{2} \text { or } x=3 \mathrm{~cm} \end{array} $$
i.e. distance of charge $q$ from $2 q$ should be $3 \mathrm{~cm}$.
Electric field at $q$
$$ E=\frac{K(2 q)}{\left(3 \times 10^{-2}\right)^{2}}-\frac{K(8 q)}{\left(6 \times 10^{-2}\right)^{2}}=0 $$
BETA
