Chapter 17, Problem 16P

To determine

The work done required.

Answer to Problem 16P

The work done required is 0.512 J.

Explanation of Solution

Given information:

Distance, $r_2=46\text{ cm}$

Charge, $q_1=q_1=35\text{μC}$

Constant, $k=9\times {10}^9{\text{ Nm}}^2{\text{/C}}^2$

Formula Used:

The formula for the electric potential is given by,

  $V=k\frac{q}{r}$

Here, k is the constant, q is the charge, r is the distance.

The formula for work is given by,

  $W=q\Delta V$

Here, W is the required work done, and $\Delta V$ is the change in electric potential.

Calculation:

For mid-way, calculate the potential.

  $\begin{array}{c}{V}_1=\frac{k{q}_1}{r_1}+\frac{k{q}_2}{r_2}\\ =k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)\end{array}$

Calculate the r₁ ,

  $\begin{array}{c}{r}_1=\frac{r_2}{2}\\ =\frac{46\text{ cm}}{2}\\ =23\text{ cm}\end{array}$

Substitute the values.

  $\begin{array}{c}{V}_1=k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)\\ =9\times {10}^9{\text{ Nm}}^2{\text{/C}}^2\left(\frac{35\mu }{23\text{ cm}}+\frac{35\mu }{\text{46 cm}}\right)\\ =2.739\times {10}^6\text{ V}\end{array}$

For 12 cm away,

  $\begin{array}{c}{r}_1=\frac{r_2}{2}-12\text{ cm}\\ =\frac{46\text{ cm}}{2}-12\text{ cm}\\ =23\text{ cm}-12\text{ cm}\\ =11\text{ cm}\end{array}$

and,

  $\begin{array}{c}{r}_2=23\text{ cm}+12\text{ cm}\\ =35\text{ cm}\end{array}$

Now, calculate the potential,

  $\begin{array}{c}{V}_2=k\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)\\ =9\times {10}^9{\text{ Nm}}^2{\text{/C}}^2\left(\frac{35\mu }{\text{11 cm}}+\frac{35\mu }{\text{35 cm}}\right)\\ =3.763\times {10}^6\text{ V}\end{array}$

Calculate change in electric potential.

  $\begin{array}{c}\Delta V=V_2-V_1\\ =\left(3.763\times {10}^6\text{ V}\right)-\left(2.739\times {10}^6\text{ V}\right)\\ =1.024\times {10}^6\text{ V}\end{array}$

Calculate the work done.

  $\begin{array}{c}W=q\Delta V\\ =\left(0.5\mu \text{C}\right)1.024\times {10}^6\text{ V}\\ =\text{0}\text{.512 J}\end{array}$

Conclusion:

Thus, the work done required is 0.512 J.