Question 827: Electrostats & Capacitance

Question

Two charged particles each having charge ' $q$ ' and mass ' $m$ ' are held at rest while their separation is ' $r$ '. The speed of the particles when their separation is ' $\frac{\mathrm{r}}{2}$ ' will be ( $\varepsilon_0=$ permittivity of the medium)

Accepted Answer

To determine the speed of two charged particles when their separation changes, we use the principle of conservation of energy. Consider the following steps:
Initially, the two charged particles, each with charge $ q $ and mass $ m $, are at rest and separated by a distance $ r $. The initial kinetic energy (K.E.) of the system is zero because the particles are not moving. The initial potential energy (P.E.) is given by the electrostatic potential energy formula:
$ 	ext{Initial P.E.} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r} $
When the particles move closer such that their separation is reduced to $ \frac{r}{2} $, they gain kinetic energy. The final kinetic energy (assuming that both particles gain the same speed $ v $) can be expressed as:
$ 	ext{Final K.E.} = 2 	imes \frac{1}{2} m v^2 = m v^2 $
The final potential energy when the separation is $ \frac{r}{2} $ is:
$ 	ext{Final P.E.} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r/2} = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{2q^2}{r} $
According to the conservation of energy principle:
$ (	ext{Initial K.E.} + 	ext{Initial P.E.}) = (	ext{Final K.E.} + 	ext{Final P.E.}) $
Substituting the expressions for the initial and final energies:
$ \frac{1}{4 \pi \varepsilon_0} \cdot \frac{q^2}{r} = m v^2 + \frac{1}{4 \pi \varepsilon_0} \cdot \frac{2q^2}{r} $
Rearranging terms to solve for $ m v^2 $:
$ m v^2 = \frac{1}{4 \pi \varepsilon_0}\left(\frac{2q^2}{r} - \frac{q^2}{r}ight) = \frac{q^2}{4 \pi \varepsilon_0} \cdot \frac{1}{r} $
Then:
$ m v^2 = \frac{q^2}{4 \pi \varepsilon_0} \cdot \frac{1}{r} $
Solving for $ v $:
$ v = \frac{q}{\sqrt{4 \pi \varepsilon_0 m r}} $
Thus, the speed of the particles when their separation is reduced to $ \frac{r}{2} $ is given by:
$ v = \frac{q}{\sqrt{4 \pi \varepsilon_0 m r}} $

Answer

$\frac{q}{4 \pi \varepsilon_0 \mathrm{mr}}$

Answer

$\frac{\mathrm{q}}{2 \pi \varepsilon_0 \mathrm{mr}}$

Answer

$\frac{\mathrm{q}}{\sqrt{4 \pi \varepsilon_0 \mathrm{mr}}}$

Answer

$\frac{\mathrm{q}^2}{4 \pi \varepsilon_0 \mathrm{mr}}$

Electrostats & Capacitance

Two charged particles each having charge ' q ' and mass ' m ' are held at rest while their separation is ' r '. The speed of the particles when their separation is ' \frac{\mathrm{r}}{2} ' will be ( \varepsilon_0= permittivity of the medium)

Choose the correct answer: