Question

My Notes An electron is released from rest on the axis of a uniform positively charged ring, 0.500 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 0.700 m, how fast will the electron be moving when it reaches the center of the ring?

Answer 1

Velocity of the electron = v = 1.2\times 10^8\ m/s.

Step-by-step explanation:

Given,

• Mass of the electron =
  `m_e\ =\ 9* 10^(-31)\ kg`
• Charge on the electron =
  `q_e\ =\ 1.62* 10^(-19)\ C`
• Charge density of the ring =
  `\rho\ =\ +1.00* 10^(-6)\ C/m`
• Radius of the ring = R = 0.70 m
• Distance between the electron ant the center or the ring = x = 0.5 m

Now total charge on the ring =
`Q\ =\ \rho* 2\pi R`

Potential energy due to the charged ring to the point on the x-axis is

`P.E.\ =\ (KQq_e)/(√(R^2\ +\ x^2))\\`

Let v be the velocity of the electron at the center of the ring.

Total kinetic energy of the electron =
`(1)/(2)m_ev^2\\`

Now, From the conservation of energy,

the total potential energy of the electron at initially is converted to the total kinetic energy of the electron at the center of the ring,

`\therefore P.E.\ =\ K.E.\\\Rightarrow (KQ)/(√(R^2\ +\ x^2))\ =\ (1)/(2)m_ev^2\\\Rightarrow v\ =\ sqrt{(2 KQq_e)/(m_e√(R^2\ +\ x^2))}\\\Rightarrow v\ =\ \sqrt{(2Kq_e\rho * 2\pi R)/(m_e√(R^2\ +\ x^2))}\\\Rightarrow v\ =\ \sqrt{(2* 9* 10^9* 1.0* 10^(-6)* 2* 3.14* 0.7* 1.6* 10^(-19))/(9* 10^(-31)* √(0.7^2\ +\ 0.5^2))}\\\Rightarrow v\ =\ 1.2* 10^8\ m/s.`

Hence the velocity of the electron on the center of the charged ring is
`1.2* 10^8\ m/s.`