Question

Two identical particles each of mass m and having charges -q and +q are revolving in a circle of radius r under the influence of electric attraction. Kinetic energy of each particle is $(k=\frac{1}{4\pi {ϵ}_0})$:

• A. $k{q}^2/4r$
• B. $k{q}^2/2r$
• C. $k{q}^2/8r$
• D. $k{q}^2/r$

Answer 1

The correct option is C $k{q}^2/8r$

The magnitude of force of attraction is $|F|=\frac{q^2}{4\pi {ϵ}_0(2r{)}^2}$.
Here the centrifugal force is balanced by the force of attraction in equilibrium.
i.e, $\frac{m{v}^2}{r}=|F|=\frac{q^2}{4\pi {ϵ}_0(2r{)}^2}=k\frac{q^2}{4r^2}$
$\Rightarrow m{v}^2=k\frac{q^2}{4r}$
Kinetic energy $=\left(1/2\right)m{v}^2=\frac{k{q}^2}{8r}$