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

Answer:

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

Explanation:

Given,

• Mass of the electron = $m_e\ =\ 9\times 10^{-31}\ kg$
• Charge on the electron = $q_e\ =\ 1.62\times 10^{-19}\ C$
• Charge density of the ring = $\rho\ =\ +1.00\times 10^{-6}\ C/m$
• 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 = $Q\ =\ \rho\times 2\pi R$

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

$P.E.\ =\ \dfrac{KQq_e}{\sqrt{R^2\ +\ x^2}}$

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

Total kinetic energy of the electron = $\dfrac{1}{2}m_ev^2$

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,

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

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