A very small sphere with positive charge $\mathit{q}\mathbf{=}\mathbf{+}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\mu }\mathit{C}$is released from rest at a point $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\text{cm}}$from a very long line of uniform linear charge density $\mathit{\lambda }\mathbf{=}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathit{\mu }\mathit{C}\mathbf{/}\mathit{m}$. What is the kinetic energy of the sphere when it is $\mathbf{4}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\text{cm}}$from the line of charge if the only force on it is the force exerted by the line of charge?

The total energy of a system is constant even though the system undergoes a change, and energy can change forms but is not lost during the process.

${\mathit{K}}_{\mathbf{a}}\mathbf{+}{\mathit{U}}_{\mathbf{a}}\mathbf{=}{\mathit{K}}_{\mathbf{b}}\mathbf{+}{\mathit{U}}_{\mathbf{b}}$

Where K is the kinetic energy and U is the potential energy.

Given data:

• Distance between the sphere and the wire (initial location), $r_a=1.5\text{cm}$.
• Distance between the sphere and the wire (final location), $r_b=4.5\text{cm}$.
• The linear charge density of the wire,$\lambda =3.0\times {10}^{-6}\text{C/m}$.
• Initial kinetic energy,$K_a=0$.
• Charge on the sphere,$q=8\times {10}^{-6}\text{C}$.

From the law of conservation,

$K_a+U_a=K_b+U_b$

Substitute the values, and we get,

$K_b=U_a-U_b$

The electric potential between $a$and $b$is given by,

$V_a-V_b=\frac{\lambda }{2\pi {\epsilon }_0}\ln \frac{r_b}{r_a}$

Substitute all the values in the above equation, and we get,

$$\begin{gathered} V_a-V_b=\frac{3.0\times {10}^{-6}}{2\pi \times 8.85\times {10}^{-12}}\ln \left(\frac{4.5}{1.5}\right) \\ =5.9\times {10}^4\text{V} \end{gathered}$$

Also, the potential between a and b is ;

$$\begin{gathered} U_a-U_b=q\left(V_a-V_b\right) \\ =8\times {10}^{-6}\times 5.9\times {10}^4 \\ =0.472\text{J} \end{gathered}$$

Now, comparing the previous notation with equation (1), and we get,

$K_b=0.472\text{J}$

Thus, the kinetic energy of the sphere will be $\mathbf{0}\mathbf{.}\mathbf{472}\mathbf{}\mathbf{\text{J}}$.