Chapter 32, Problem 48A

Two pellets, each with a charge of $1\mu C$, are separated by a distance of 0.30 m. Show that the electric force between them is 0.1 N.

To determine

To Show: Electric force between the two forces is 0.1N

Explanation of Solution

Given:

Two pellets have equal charge of 1µC.

They are separated by a distance of 0.30 m.

Formula used:

The force is given as,

  $F_1=k\frac{q_1{q}_2}{{(r)}^2}$

Here,q₁and q₂are two charges and r is the separation between them.

Calculation:

Given $q_1=q_2=1\text{μC}$ and r=0.30 m

From the coulomb’s law

  $F_1=k\frac{q_1{q}_2}{{(r)}^2}$

Here k is constant and value of k is $9.0\times {10}^9 {\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}$

Substituting the values,

  $\begin{array}{c}{F}_1=\frac{9\times {10}^9\times 1\times {10}^{-6}\times 1\times {10}^{-6}}{0.30\times 0.30}\\ =\frac{9\times {10}^{-3}}{0.09}\\ =\frac{1}{10}\\ =0.1\text{N}\end{array}$

Conclusion:

Therefore, the force between the charges is 0.1N.