Question

Question: A charge of -9.0 nCis uniformly distributed around a thin plastic ring lying in a y-zplane with the ring center at the origin. A-6.0 pC particle is located on the xaxis at x = 3.0 m. For a ring radius of, how much work must an external force do on the particle to move it to the origin?

Short answer

Answer:

The work done by an external force do on the particle to move it to the origin is $1.8\times {10}^{-10}\text{J}$.

Step-by-step solution

The given data

1. Radius of the ring, r = 1.5 m
2. Distributed charge on the ring,$q_2=-6.0\text{pC}$
3. Charge at the point P,$q_2=-6.0\text{pC}$
4. Distance from the origin to the point P,$x=3.0\text{m}$

Understanding the concept of work done  

Using the concept of potential energy at a distance r and the potential energy at an axial point, we can get the value of the required external work by calculating the difference between the energies.

Formulae:

Potential energy of a system of two charges is given by the relation,$U=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q_1{q}_2}{r}\right)$(i)

Work done can obtained by using the expression$W=\Delta V$,(ii)

Calculation of the work done  

The required of the work done to bring to its origin is given using equation (i) in equation (ii) as follows:

$$\begin{gathered} W=\frac{1}{4\pi {\epsilon }_0}{q}_1{q}_2\left(\frac{1}{r}-\frac{1}{\sqrt{r^2+x^2}}\right) \\ =\left(9.0\times {10}^9\right)\left(9.0\times {10}^{-9}\text{C}\right)\left(-6.0\times {10}^{-12}\text{C}\right)\left[\frac{1}{1.5\text{m}}-\frac{1}{\sqrt{{(1.5\text{m})}^2+{(3.0\text{m})}^2}}\right] \\ =1.79\times {10}^{-10}\text{J} \\ =1.8\times {10}^{-10}\text{J} \end{gathered}$$

Hence, the value of the work is $1.8\times {10}^{-10}\text{J}$.