Chapter 6, Problem 17E

(a) A metal sphere with a charge of +1 × 10⁻⁵ C is 10 cm from another metal sphere with a charge of −2 × 10⁻⁵ C. Find the magnitude of the attractive force acting on each sphere. (b) The two spheres are brought in contact and again separated by 10 cm. Find the magnitude of the new force acting on each sphere.

To determine

The magnitude of the attractive force acting on each sphere.

Answer to Problem 17E

The magnitude of the attractive force acting on each sphere is 180 N.

Explanation of Solution

Given data:

$\begin{array}{l}{Q}_1=1\times {10}^{-5}\text{C}\\ Q_2=-2\times {10}^{-5}\text{C}\\ R=10\text{cm}\end{array}$

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 charged particle,

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

$K$ is the coulomb’s constant, and

$R$ is the distance between the two charged particles.

Substitute $1\times {10}^{-5}\text{C}$ for $Q_1$, $-2\times {10}^{-5}\text{C}$ for $Q_2$, $10\text{cm}$ for $R$ and $9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}$ for $K$ in equation (1),

$\begin{array}{c}F=\left(9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\right)\frac{\left(1\times {10}^{-5}\text{C}\right)\left(-2\times {10}^{-5}\text{C}\right)}{{\left(10\text{cm}\right)}^2}\\ =\left(9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\right)\frac{\left(1\times {10}^{-5}\text{C}\right)\left(-2\times {10}^{-5}\text{C}\right)}{{\left(0.10\text{m}\right)}^2}\left\{\because 1\text{m}=100\text{cm}\right\}\\ =\left(9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\right)\frac{\left(1\times {10}^{-5}\text{C}\right)\left(-2\times {10}^{-5}\text{C}\right)}{\left(0.01{\text{m}}^2\right)}\\ =-180\text{N}\end{array}$

So, the magnitude of the attractive force on each sphere is,

$\begin{array}{c}\left|F\right|=\left|-180\text{N}\right|\\ =180\text{N}\end{array}$

Conclusion:

Hence, the magnitude of the attractive force acting on each sphere is 180 N.

To determine

The magnitude of the new force acting on each sphere when the 2 spheres are brought in contact and again separated by 10 cm.

Answer to Problem 17E

The magnitude of the new force acting on each sphere, when the two spheres are brought in contact and separated by 10 cm is 22.5 N.

Explanation of Solution

When the two charged metal spheres are brought in contact, the charges in the 2 metal spheres gets equally divided such that, $Q_1=Q_2=Q$.

So the charge of $Q_1$ and $Q_2$ after the contact is,

$\begin{array}{c}Q=\frac{Q_1+Q_2}{2}\\ =\frac{1\times {10}^{-5}\text{C}+\left(-2\times {10}^{-5}\text{C}\right)}{2}\\ =\frac{1\times {10}^{-5}\text{C}-2\times {10}^{-5}\text{C}}{2}\\ =\frac{-1\times {10}^{-5}\text{C}}{2}\end{array}$

$Q=-5\times {10}^{-6}\text{C}$

Substitute $-5\times {10}^{-6}\text{C}$ for $Q_1$ and $Q_2$, $9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}$ for $K$ and 10 cm for $R$ in equation (1),

$\begin{array}{c}F=\left(9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\right)\frac{\left(-5\times {10}^{-6}\text{C}\right)\left(-5\times {10}^{-6}\text{C}\right)}{{\left(10\text{cm}\right)}^2}\\ =\left(9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\right)\frac{\left(-5\times {10}^{-6}\text{C}\right)\left(-5\times {10}^{-6}\text{C}\right)}{{\left(\text{0}\text{.1m}\right)}^2}\left\{\because 1\text{m}=100\text{cm}\right\}\\ =\left(9\times {10}^9\frac{\text{N}\cdot {\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\right)\frac{\left(-5\times {10}^{-6}\text{C}\right)\left(-5\times {10}^{-6}\text{C}\right)}{\left(\text{0}{\text{.01m}}^2\right)}\\ =22.5\text{N}\end{array}$

Conclusion:

Hence, the magnitude of the new force acting on each sphere when the two spheres are brought in contact and separated by 10 cm is 22.5 N.