Question

A uniformly charged, thin ring has radius 15.0 cm and total charge +24.0 nC. An electron is placed on the ring’s axis a distance 30.0 cm from the center of the ring and is constrained to stay on the axis of the ring. The electron is then released from rest. (a) Describe the subsequent motion of the electron. (b) Find the speed of the electron when it reaches the center of the ring.

Short answer

a) At point (a), the electron will begin to move along the axis from point (a) to point (b), and the magnitude of the electric field will change as it moves, causing instantaneous force and acceleration to change as well. At point (b), the electron's velocity will reach its maximum value (b).

b) The speed of the electron when it reaches the center of the ring is$1.67x{10}^{7}mls$

Step-by-step solution

Kinetic energy

It is defined as thework needed to accelerate a body of a given mass from rest to its statedvelocity.

$K=\frac{1}{2}m{v}^2$

The subsequent motion of the electron

a) At point (a), the electron will begin to move along the axis from point (a) to point (b), and the magnitude of the electric field will change as it moves, causing instantaneous force and acceleration to change as well. At point (b), the electron's velocity will reach its maximum value (b).

The speed of the electron when it reaches the center of the ring

b) Given;

$\begin{array}{l}{\mathrm{x}}_{\mathrm{a}}=0.3\mathrm{m}\\ {\mathrm{x}}_{\mathrm{b}}=0\\ \mathrm{q}=1.6\times {10}^{-19}\mathrm{C}\\ \mathrm{a}=24.0\mathrm{nC}\\ \mathrm{a}=0.15\mathrm{m}\end{array}$

The kinetic and potential energy;

${\mathrm{U}}_a+K_a=U_b+K_b$

Initial Potential Energy;

The potential at point (a);

$V_a=\frac{kQ}{\sqrt{x_a^2+a^2}}$

Putting the values;

$\begin{array}{l}{U}_4=6V_a\\ =\frac{9\times {10}^9\times (-19\times 10-19\times 24\times 10-9}{\left(\right\{{015}^2+({003}^2}\\ =-103\times 1016]\end{array}$

Final Potential energy;

$V_b=\frac{kQ}{Q}\sqrt{x_b^2+a^2}$

Putting the values;

$\begin{array}{l}{u}_b=6V_b\\ =\frac{9\times {10}^9\times (-19\times 10-19\times 24\times 10-9}{\sqrt{(a{15}^2}+(0^2}\\ =-2.3\times 00-6]\end{array}$

By substituting;

$$\begin{gathered} -103\times {10}^{-16}=-2.3\times {10}^{-16}+\frac{m_e^{}{V^2}_b}{2} \\ {V}_b=\sqrt{\frac{2\times 1.27\times {10}^{-16}}{9.1\times {10}^{-31}}}=167\times {10}^{-7}mls \end{gathered}$$

Hence, the speed of the electron when it reaches the center of the ring is$167\times {10}^{-7}mls$