Chapter 20, Problem 45A

To determine

The new distance between the spheres A and B to have the same deflection that they had before.

Answer to Problem 45A

New distance should be reduced by $\sqrt{3}$ times to have the same deflection.

Explanation of Solution

Given:

Initially the charges on sphere A and B are same, $q_A=q_B$

Initial distance between the spheres is $r$ .

Then, the charge on B is made one-third of the charge on A, $Q_B=\frac{q_A}{3}$

Formula used:

The electrostatic force between two charges, $q_1$ and $q_2$ , $r$ distance apart is given by,

  $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$

Where ${\epsilon }_0=\text{ permittivity of free space}=8.85\times {10}^{-12}\text{F/m}$

Calculation:

The initial force between the charges will be,

  $\begin{array}{c}F=\frac{1}{4\pi {\epsilon }_0}\frac{q_A{q}_B}{r^2}\\ =\frac{1}{4\pi {\epsilon }_0}\frac{q_A^2}{r^2}\end{array}$

Now, the charge on sphere B is made one-third of the charge on sphere A, then new distance will be,

  $\begin{array}{c}F=\frac{1}{4\pi {\epsilon }_0}\frac{q_A{Q}_B}{{\left(r\text{'}\right)}^2}\\ \frac{1}{4\pi {\epsilon }_0}\frac{q_A^2}{r^2}=\frac{1}{4\pi {\epsilon }_0}\frac{\frac{q_A^2}{3}}{{\left(r\text{'}\right)}^2}\\ r\text{'}=\frac{r}{\sqrt{3}}\end{array}$

Conclusion:

Hence, the new distance should be the initial distance divided by $\sqrt{3}$ .