Chapter 8, Problem 8.1E

What is the charge on a small sphere that is attracted to another sphere having charge $\text{1}\text{.00 C}$ if the spheres are $\text{100}\text{.0 m}$ apart and the force of attraction is $\text{0}\text{.0225 N}$?

Interpretation Introduction

Interpretation:

The charge on a small sphere that is attracted to another sphere is to be calculated.

Concept introduction:

The force of attraction between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula for determining force of attraction is given below.

$F=\frac{q_1{q}_2}{4{\text{πε}}_{\text{o}}{r}^2}$

In the above equation, $F$ is the force of attraction between two charges, $q_1$ and $q_2$ are the charges, $r$ is the distance between charges and ${\text{ε}}_{\text{o}}$ is the constant. The value of ${\text{ε}}_{\text{o}}$ is defined as $8.854\times {10}^{-12}{\text{ C}}^{\text{2}}{\text{ J}}^{-\text{1}}{\text{m}}^{-\text{1}}$.

Answer to Problem 8.1E

The charge on a small sphere that is attracted to another sphere is $2.50\times {10}^{-8}\text{ C}$.

Explanation of Solution

For calculation of charge on one sphere, the above equation can be written as given below.

$q_1=\frac{4{\text{πε}}_{\text{o}}{r}^{2}F}{q_2}$

On substitution of the values in above equation, charge on one sphere obtained is given below.

$\begin{array}{rll}{q}_1& =& \frac{4{\text{πε}}_{\text{o}}{r}^{2}F}{q_2}\\ & =& \frac{4\times 3.14\times 8.854\times {10}^{-12}\times {\left(100\right)}^2\times 0.0225}{1.00}\\ & =& 2.50\times {10}^{-8}\text{ C}\end{array}$

Conclusion

The charge calculated on one sphere that is attracted to another sphere is $2.50\times {10}^{-8}\text{ C}$.