Question

A ring of diameter $\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{cm}}$is fixed in plane and carries a charge of $\mathbf{+}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathit{\mu }\mathit{C}$uniformly spread over its circumference.

(a) How much work does it take to move a tiny $\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathit{\mu }\mathit{C}$charged ball of mass $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\text{g}}$from very far away to the center of the ring?

(b) Is it necessary to take a path along the axis of the ring? Why?

(c) If the ball is slightly displaced from the center of the ring, what will it do, what is the maximum speed it will reach?

Short answer

1. The work done is$\mathbf{3}\mathbf{.}\mathbf{38}\mathbf{}\mathbf{\text{J}}$
2. No, it is not. All the charge in a ring of charge is the same distance from any point on the ring axis, so any path can be taken along the axis to calculate the work done.
3. The maximum speed will be$\mathbf{67}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{m/s}}$

Step-by-step solution

Formulas required to solve the question

Potential due to a single point charge:

$V_{ring}=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}$

Work done due to the electric field between the two charges:

$$\begin{gathered} W=\Delta U \\ =q\Delta V \\ =q\left(V_{ring}-V_{far\_away}\right) \end{gathered}$$

Total mechanical energy is conserved:

$K_1+U_1=K_2+U_2$

Where Q is the charge, r is the radius, q is the charge on a tiny ball, K is the kinetic energy, and U is potential energy.

Calculate the work done

(a)

As the charged ball is very far away, so consider the ring to be a point charge, where the potential due to a single point charge is given by:

$V_{ring}=\frac{1}{4\pi {\epsilon }_0}\frac{Q}{r}$

Where r is the radius of the ring $r=R$, so the work done due to the electric field between the two charges is given by,

$$\begin{gathered} W=\Delta U \\ =q\Delta V \\ =q\left(V_{ring}-V_{far\_away}\right) \end{gathered}$$

Where q is the charge of the ball and potential at an infinite point equals zero, so $V_{far\_away}=0$and the equation will be,

$$\begin{gathered} W=q{V}_{ring} \\ =\frac{1}{4\pi {\epsilon }_0}\frac{qQ}{R} \end{gathered}$$

Plug the values in the above equation, and we get,

$$\begin{gathered} W=(9.0\times {10}^9\text{N}·{\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})\times \frac{\left(3.00\times {10}^{-6}\text{C}\right)\left(5.00\times {10}^{-6}\text{C}\right)}{\left(0.04\text{m}\right)} \\ =3.38\text{J} \end{gathered}$$

Thus, the work done is$\mathbf{3}\mathbf{.}\mathbf{38}\mathbf{}\mathbf{\text{J.}}$

Reason for taking the path along the axis of a ring

(b)

The answer is no; it is not. All the charge in a ring of charge is the same distance from any point on the ring axis, so any path can be taken along the axis to calculate the work done.

Thus, it is not necessary to take a path along the axis of the ring.

Determine the maximum speed the ball will reach

(c)

Here, due to the same sign of charge for the ring and the ball, there will be a repulsive force, and the ball will move outward from the ring. Therefore, there will be two instants, first when the ball is close to the ring and second when the ball is far away where the total mechanical energy is conservative.

$K_1+U_1=K_2+U_2$

Where $K_1$at the center of the ring is zero and $U_2$at infinite equals zero. While $U_1$is the maximum potential and is given as:

$$\begin{gathered} U_1=\Delta U-U_2 \\ {U}_1=\Delta U-0 \\ {U}_1=W \\ {U}_1=3.38\text{J} \end{gathered}$$

Now, the relation between maximum speed and the potential is given as:

$$\begin{gathered} K_1+U_1=K_2+U_2 \\ {U}_1=K_2 \\ {U}_1=\frac{1}{2}m{v}^2 \\ v=\sqrt{\frac{2U_1}{m}} \end{gathered}$$

Plug the values, and we get,

$$\begin{gathered} v=\sqrt{\frac{2\times 3.38\text{J}}{0.0015\text{kg}}} \\ =67.0\text{m/s} \end{gathered}$$

Thus, the maximum speed will be $67.0\text{m/s}$.