Question

A ring of radius R carries a uniformly distributed charge $+Q$. A point charge q is placed on the axis of the ring and released from rest. The force experienced by the particle varies with distance from the centre as

• A. $x/(R^2+x^2{)}^{3/2}$
• B. $1/\sqrt{x}$
• C. $1/x^3$
• D. $1/x^2$

Answer 1

The correct option is A $x/(R^2+x^2{)}^{3/2}$

The electric field at a distance $x$ where charge q is placed due to ring is $E=\frac{1}{4\pi {ϵ}_0}\frac{Qx}{(x^2+R^2{)}^{3/2}}$

Now the force on charge q is $F=qE=\frac{q}{4\pi {ϵ}_0}\frac{Qx}{(x^2+R^2{)}^{3/2}}$

$\therefore F\propto \frac{x}{(x^2+R^2{)}^{3/2}}$