Question

An electron is released from rest on the axis of a uniform positively charged ring, 0.200 m from the ring's center. If the linear charge density of the ring is 0.100 µC/m and the radius of the ring is 0.400 m, how fast will the electron be moving when it reaches the center of the ring?

Answer 1

Answer:

Velocity of the electron at the centre of the ring, $v=1.37\times10^7\ \rm m/s$

Explanation:

Given:

• Linear charge density of the ring=$0.1\ \rm \mu C/m$
• Radius of the ring R=0.2 m
• Distance of point from the centre of the ring=x=0.2 m

Total charge of the ring

$Q=0.1\times2\pi R\\Q=0.1\times2\pi 0.4\\Q=0.251\ \rm \mu C$

Potential due the ring at a distance x from the centre of the rings is given by

$V=\dfrac{kQ}{\sqrt{(R^2+x^2)}}$

The potential difference when the electron moves from x=0.2 m to the centre of the ring is given by

$\Delta V=\dfrac{kQ}{R}-\dfrac{kQ}{\sqrt{(R^2+x^2)}}\\\Delta V={9\times10^9\times0.251\times10^{-6}} \left( \dfrac{1}{0.4}-\dfrac{1}{\sqrt{(0.4^2+0.2^2)}} \right )\\\Delta V=5.12\times10^2\ \rm V$

Let$\Delta U$ be the change in potential Energy given by

$\Delta U=e\times \Delta V\\\Delta U=1.67\times10^{-19}\times5.12\times10^{2}\\\Delta U=8.55\times10^{-17}\ \rm J$

Change in Potential Energy of the electron will be equal to the change in kinetic Energy of the electron

$\Delta U=\dfrac{mv^2}{2}\\8.55\times10^{-17}=\dfrac{9.1\times10^{-31}v^2}{2}\\v=1.37\times10^7\ \rm m/s$

So the electron will be moving with $v=1.37\times10^7\ \rm m/s$

Answer 2

The change of potential energy of electron will be equal to the change in kinetic energy.

The electron be moving when it reaches the center of the ring with the speed of $1.37\times10^{7}\rm m/s$.

What is electric potential energy?

Electric potential energy is the energy which is required to move a unit charge from a point to another point in the electric field.

It can be given as,

$U=q\Delta V\\U=qk\lambda A[\dfrac{1}{R}- \dfrac{1}{\sqrt{R^2+x^2}} ]$

Here, $q$ is the charge, $\lambda$ is the linear charge density, $A$ is the area, $R$ is the radius and $x$ is the distance.

Given information-

The distance of the point from the center of the ring is 0.200 meters.

The linear charge  density of the ring is 0.100 µC/m.

The radius of the ring is 0.400 meters.

Put the values in the above formula as,

$\Delta U=1.67\times10^{-19}\times 9\times10^9 \times0.1\times2\pi0.4[\dfrac{1}{0.4}- \dfrac{1}{\sqrt{0.4^2+0.2^2}} ]\\\Delta U=8.55\times10^{-17} \rm J$

AS the change of potential energy of electron will be equal to the change in kinetic energy. Thus,

$\dfrac{mv^2}{2} =\Delta U\\\dfrac{9.1\times10^{-17}v^2}{2} =8.55\times10^{-17}\\v=1.37\times10^7$

Thus the electron be moving when it reaches the center of the ring with the speed of $1.37\times10^{7}\rm m/s$.

Learn more about the electric potential energy here;

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