Chapter 16, Problem 1P

To determine

The magnitude of the force between two similar charges that are $9.3\text{cm}$ apart.

Answer to Problem 1P

The magnitude of the force $13.46\text{N}$ .

Explanation of Solution

Given information:

Magnitude of charges, $Q_1=Q_2=3.60\text{μC}=3.60\times {10}^{-6}\text{C}$

Distance between charges, $r=9.3\text{cm}=\text{9}\text{.3}\times {\text{10}}^{-2}\text{m}$

Formula used:

Write the expression to calculate the magnitude of force acting between two charges using Coulomb’s law.

  $F=k\frac{Q_1{Q}_2}{r^2}$...........(1)

Where,

• The force acting between two charges is $F$ .
• The Coulomb’s constant is $k$ and its value is $8.988\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^2$ .

Calculation:

Substitute the values in equation (1).

  $\begin{array}{c}F=\left(8.988\times {10}^9\right)\frac{\left(3.60\times {10}^{-6}\right)\left(3.60\times {10}^{-6}\right)}{{\left(\text{9}\text{.3}\times {\text{10}}^{-2}\right)}^2}\\ =8.988\times {10}^9\times 1.498\times {10}^{-9}\\ =13.46\text{N}\end{array}$

Therefore, themagnitude of the force $13.46\text{N}$ .