Question

A charge $+Q$ is uniformly distributed over a thin ring with radius $R$. A negative point charge $-Q$ and mass $m$ starts from at a point far away from the centre of the ring and moves towards the centre. Find the velocity of this particle at the moment it passes through the centre of the ring.

Answer 1

The figure shows the situation where the point charge -Q is initially.

The force of attraction between ring with the charge +Q and point charge -Q is given by F = QE.

Let us find the electric field due to charge +Q at a distance x from the center of the ring.

For a very small element of the ring with charge dQ, the field at point P is

$\text{dE}=\frac{dQ}{4\pi {ϵ}_0(AP{)}^2}$

As the particle start moving toward the center of the ring from this distance we can write for the field

$\text{dE}\sin \theta =\frac{dQ}{4\pi {ϵ}_0(AP{)}^2}\cdot \frac{AO}{AP}=\frac{R\cdot dQ}{4\pi {ϵ}_0(R^2+x^2{)}^{3/2}}$

Thus the net field at the point P is $E=\int dE\sin \theta =\frac{R\cdot Q}{4\pi {ϵ}_0(R^2+x^2)}$

Thus total force of attraction is given by $\text{F}=\frac{R\cdot Q^2}{4\pi {ϵ}_0(R^2+x^2{)}^{3/2}}...\left(i\right)$

If the particle moves with velocity $v_0$ then due to coulomb's force of attraction the particle starts accelerating.

Thus its speed increases by amount $v_0+at...\left(ii\right)$

The acceleration of the particle is given by $a=\frac{F}{m}=\frac{R\cdot Q^2}{4m\pi {ϵ}_0(R^2+x^2{)}^{3/2}}$

When point charge reaches center of the circle x=0 and thus we get from (ii)

$v=v_0+\frac{Q^2}{4m\pi {ϵ}_0{R}^2}$