91Ó°ÊÓ

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

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

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) Given;

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

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;

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

Final Potential energy;

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

Putting the values;

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

By substituting;

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

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