Question

Three-point charges q, 2q and 8q are to be placed on a 9 cm long straight line. Find 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 charge q due to the other two charges?

Answer 1

Given $q$, $2q$, $8q$ three charges present,now the system potential energy to be minimum we will place $q$ charge in between $2q$ & $8q$.

Such that,

Now, The potential energy of the system is given by,

$u=\frac{K{q}_1{q}_2}{r_{12}}+\frac{K{q}_2{q}_3}{r_{23}}+\frac{K{q}_3{q}_1}{r_{31}}$

$u=K[\frac{{2q}^2}{x}+\frac{{8q}^2}{9-x}+\frac{{16q}^2}{9}]$

$\Rightarrow u=K{q}^2[\frac{2}{x}+\frac{8}{9-x}+\frac{16}{9}]$

Now potential energy of the system depends on the value of $x$.

$\therefore u^1=\frac{2}{x}+\frac{8}{9-x}$ {u : minimumizing differentiating w.r.t $x$}

will give $\frac{{du}^1}{dx}=\frac{-2}{x^2}+\frac{8}{{(9-x)}^2}$

for $u^1$ to be minimum,

$\frac{{du}^1}{dx}=0\Rightarrow \frac{-2}{x^2}+\frac{8}{{(9-x)}^2}=0$

$\frac{1}{x^2}=\frac{4}{{(9-x)}^2}$

$\Rightarrow$ $x=3cm$ or $-9cm$

Now, $x=-9cm$ is not applicable hence we will take $x=3cm$

Potential energy at $x=3$,

${u/}_{x=3}={Kq}^2[\frac{2}{3}+\frac{8}{6}+\frac{16}{9}]=\frac{{Kq}^2}{3}[6+\frac{16}{3}]=\frac{34K{q}^2}{9}U$

The minimum value of $P.E=\frac{34K{q}^2}{9}$

In this situation electric field at charge $q$ is given by,

$\overrightarrow{E}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2$

where,

${\overrightarrow{E}}_1=\frac{K2q}{{\left(3\right)}^2}\&{\overrightarrow{E}}_2=\frac{K8q}{{\left(6\right)}^2}$

$=\frac{2Kq}{9}\left(\hat{i}\right)$ $=\frac{2Kq}{9}(-i)$

as ${\overrightarrow{E}}_1$ & ${\overrightarrow{E}}_2$ are opposite directed resultant is given by,

$\overrightarrow{E}=\frac{2Kq}{9}-\frac{2Kq}{9}=0$