Chapter 6, Problem 16E

Two small spheres are given identical positive charges. When they are 1 cm apart, the repulsive force acting on each of them is 0.002 N. What would the force be if (a) the distance is increased to 3 cm? (b) one charge is doubled? (c) both charges are tripled? (d) one charge is doubled and the distance is increased to 2 cm?

To determine

The force between two small spheres if the distance is increased to 3 cm.

Answer to Problem 16E

The force if the distance is increased to 3 cm is 0.00022 N.

Explanation of Solution

Given data:

$\begin{array}{l}F=0.002\text{N}\\ R=1\text{cm}\end{array}$

Formula used:

Consider the expression for the electrostatic force between two charges.

$F=K\frac{Q_{\text{1}}{Q}_{\text{2}}}{R^2}$ (1)

Here,

$Q_1$ is the charge of the first charge particle,

$Q_2$ is the charge of the second charge particle,

$K$ is the coulomb’s constant, and

$R$ is the distance between the two electrons.

Substitute $0.002\text{N}$ for $F$, $1\text{cm}$ for $R$ and $9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}$ for $K$ in equation (1),

$\begin{array}{l}0.002\text{N}=9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{Q_{\text{1}}{Q}_{\text{2}}}{{\left(1\text{cm}\right)}^2}\\ 0.002\text{N}=9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{Q_{\text{1}}{Q}_{\text{2}}}{{\left(0.01\text{m}\right)}^2}\left\{\because 1\text{m}=100\text{cm}\right\}\end{array}$

Solve the equation to find $Q_1{Q}_2$.

$Q_{\text{1}}{Q}_{\text{2}}=\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}$

Substitute $\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}$ for $Q_{\text{1}}{Q}_{\text{2}}$, $3\text{cm}$ for $R$ and $9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}$ for $K$ in equation (1),

$\begin{array}{c}F=9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}}{{\left(3\text{cm}\right)}^2}\\ =9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}{\left(0.03\text{m}\right)}^2}\left\{\because 1\text{m}=100\text{cm}\right\}\\ =0.002\text{N}\frac{{\left(0.01\text{m}\right)}^2}{{\left(0.03\text{m}\right)}^2}\\ =0.002\text{N}\frac{1\times {10}^{-4}{\text{m}}^{\text{2}}}{9\times {10}^{-4}{m}^2}\end{array}$

$\begin{array}{c}F=2.22\times {10}^{-4}\text{N}\\ =0.00022\text{N}\end{array}$

Conclusion:

Hence, the force if the distance is increased to 3 cm is 0.00022 N.

To determine

The force between 2 small spheres if one of the charge is doubled.

Answer to Problem 16E

The force between 2 small spheres if one of the charge is doubled is 0.004 N.

Explanation of Solution

Substitute $0.002\text{N}$ for $F$, $1\text{cm}$ for $R$ and $9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}$ for $K$ in equation (1),

$\begin{array}{l}0.002\text{N}=9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{Q_{\text{1}}{Q}_{\text{2}}}{{\left(1\text{cm}\right)}^2}\\ 0.002\text{N}=9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{Q_{\text{1}}{Q}_{\text{2}}}{{\left(0.01\text{m}\right)}^2}\left\{\because 1\text{m}=100\text{cm}\right\}\end{array}$

Solve the equation to find $Q_1{Q}_2$.

$Q_{\text{1}}{Q}_{\text{2}}=\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}$

Since one of the charge is doubled, multiply 2 on both sides.

$2Q_{\text{1}}{Q}_{\text{2}}=2\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}$

Substitute $2\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}$ for $Q_{\text{1}}{Q}_{\text{2}}$, $1\text{cm}$ for $R$ and $9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}$ for $K$ in equation (1),

$\begin{array}{c}F=9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{2\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}}{{\left(1\text{cm}\right)}^2}\\ =9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{2\left(0.002\text{N}\right){\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}{\left(0.01\text{m}\right)}^2}\left\{\because 1\text{m}=100\text{cm}\right\}\\ =2\left(0.002\text{N}\right)\\ =0.004\text{N}\end{array}$

Conclusion:

Hence, the force between two small spheres if one of the charge is doubled is 0.004 N.

To determine

The force between 2 small spheres if both of the charges are tripled.

Answer to Problem 16E

The force between 2 small spheres if both of the charges are tripled is 0.018 N.

Explanation of Solution

Substitute $0.002\text{N}$ for $F$, $1\text{cm}$ for $R$ and $9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}$ for $K$ in equation (1),

$\begin{array}{l}0.002\text{N}=9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{Q_{\text{1}}{Q}_{\text{2}}}{{\left(1\text{cm}\right)}^2}\\ 0.002\text{N}=9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{Q_{\text{1}}{Q}_{\text{2}}}{{\left(0.01\text{m}\right)}^2}\left\{\because 1\text{m}=100\text{cm}\right\}\end{array}$

Solve the equation to find $Q_1{Q}_2$.

$Q_{\text{1}}{Q}_{\text{2}}=\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}$

Since both of the charges are tripled, multiply $3^2$ on both sides.

$3^2{Q}_{\text{1}}{Q}_{\text{2}}=3^2\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}$

Substitute $3^2\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}$ for $Q_{\text{1}}{Q}_{\text{2}}$, $1\text{cm}$ for $R$ and $9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}$ for $K$ in equation (1),

$\begin{array}{c}F=9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{3^2\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}}{{\left(1\text{cm}\right)}^2}\\ =9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{3^2\left(0.002\text{N}\right){\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}{\left(0.01\text{m}\right)}^2}\left\{\because 1\text{m}=100\text{cm}\right\}\\ =3^2\left(0.002\text{N}\right)\\ =9\left(0.002\text{N}\right)\end{array}$

$F=0.018\text{N}$

Conclusion:

Hence, the force between 2 small spheres if both of the charges are tripled is 0.018 N.

To determine

The force between 2 small spheres if one charge is doubled and the distance is increased to 2 cm.

Answer to Problem 16E

The force between 2 small spheres if one charge is doubled and the distance is increased to 2 cm is 0.001 N.

Explanation of Solution

Substitute $0.002\text{N}$ for $F$, $1\text{cm}$ for $R$ and $9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}$ for $K$ in equation (1),

$\begin{array}{l}0.002\text{N}=9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{Q_{\text{1}}{Q}_{\text{2}}}{{\left(1\text{cm}\right)}^2}\\ 0.002\text{N}=9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{Q_{\text{1}}{Q}_{\text{2}}}{{\left(0.01\text{m}\right)}^2}\left\{\because 1\text{m}=100\text{cm}\right\}\end{array}$

Solve the equation to find $Q_1{Q}_2$.

$Q_{\text{1}}{Q}_{\text{2}}=\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}$

Since one of the charge is doubled, multiply 2 on both sides.

$2Q_{\text{1}}{Q}_{\text{2}}=2\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}$

Substitute $2\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}$ for $Q_{\text{1}}{Q}_{\text{2}}$, $2\text{cm}$ for $R$ and $9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}$ for $K$ in equation (1),

$\begin{array}{c}\frac{F}{}=9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{2\frac{0.002\text{N}{\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}}}{{\left(2\text{cm}\right)}^2}\\ =9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\frac{2\left(0.002\text{N}\right){\left(0.01\text{m}\right)}^2}{9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}{\left(0.02\text{m}\right)}^2}\left\{\because 1\text{m}=100\text{cm}\right\}\\ =2\left(0.002\text{N}\right)\frac{{\left(0.01\text{m}\right)}^2}{{\left(0.02\text{m}\right)}^2}\\ =2\left(0.002\text{N}\right)\frac{\left(0.0001{\text{m}}^2\right)}{\left(0.0004{\text{m}}^2\right)}\end{array}$

$F=0.001\text{N}$

Conclusion:

Hence, the force between two small spheres if one charge is doubled and the distance is increased to 2 cm is 0.001 N.