The charges $+4 q$ and $+q$ are placed at the ends of a straight line. A third charge $-q$ is placed

Question

The charges $+4 q$ and $+q$ are placed at the ends of a straight line. A third charge $-q$ is placed at the mid point of this line. Then for displacements parallel to the charges the charge $+q$ will\

Professor PC Thomas Classes · Answer

Force on $+q$ by $4 q=\frac{A(4 q) q}{(2 r)^{2}}=\frac{A q^{2}}{r^{2}}$\ Force on $+q$ by $-q=\frac{A(q)(-q)}{r^{2}}=\frac{-A q^{2}}{r^{2}}$ Thus the two forces are equal and opposite. Therefore the charge $q$ will be in equilibrium. To know whether it is in stable equilibrium or not displace $(+q)$ slightly to the right by $x$. Then force on $+q$ to the right by ( $+4 q$ ).\ $=\frac{A(4 q)(q)}{(2 r+x)^{2}}=\frac{A 4 q^{2}}{(2 r+x)^{2}}$\ Force on (+q) to the left by $(-\mathrm{q})$ is $=\frac{\mathrm{A}(\mathrm{q})(-\mathrm{q})}{(\mathrm{r}+\mathrm{x})^{2}}$\ $=\frac{A q^{2}}{(r+x)^{2}}$ numerically.\ $\frac{	ext { Force to theright }}{	ext { Force to theleft }}=\frac{\mathrm{A} 4 \mathrm{q}^{2}}{(2 \mathrm{r}+\mathrm{x})^{2}} \frac{(\mathrm{r}+\mathrm{x})^{2}}{\mathrm{Aq}^{2}}$\ $=\frac{4(\mathrm{r}+\mathrm{x})^{2}}{(2 \mathrm{r}+\mathrm{x})}=\frac{4(\mathrm{r}+\mathrm{x})^{2}}{\mathrm{r}(\mathrm{r}+\mathrm{x} / 2)^{2}}$\ This is greater than 1. Therefore force to the right is greater. Hence the body is in unstable equilibrium.\