Question

The resultant force acting on the charged bead is

• A. $F=-\frac{k2Qq}{x_0^2+R^2}\frac{x_0}{r}$
• B. $F=\frac{k2Qq}{x_0^2+R^2}\frac{x_0}{r}$
• C. $F=-\frac{k2Qq}{x_0^2-R^2}\frac{x_0}{r}$
• D. $F=\frac{k2Qq}{x_0^2-R^2}\frac{x_0}{r}$

Answer 1

Answer: (A)

$F=-\frac{k2Qq}{x_0^2+R^2}\frac{x_0}{r}$

Here the forces on $-q$ will be same due to charges because they are equal. Since its sine component cancel out.
The electric filed at $-q$ due to $Q$ is $E=k\frac{Q}{r^2}$.
Net force on $-q$ is $F_{net}=2F\cos \theta =2(-q)E.\frac{x_0}{r}=-2qEk\frac{Q}{r^2}.\frac{x_0}{r}=-\frac{2qkQ}{x_0^2+R^2}.\frac{x_0}{r}$