A glass bead charged to +3.5 nC exerts an 8.0 x 10^-4 N repulsive electric force on a plastic bead 2.9 cm away. What is the charge on the plastic bead?

Steps to Solve

1. Identify the Given Values

We know:

• Charge on the glass bead, $q_1 = +3.5 , \text{nC} = 3.5 \times 10^{-9} , \text{C}$.
• Electric force, $F = 8.0 \times 10^{-4} , \text{N}$.
• Distance between beads, $r = 2.9 , \text{cm} = 0.029 , \text{m}$.

2. Use Coulomb's Law

Coulomb's law states:

$$ F = k \frac{|q_1 q_2|}{r^2} $$

Where:

• $F$ is the force between the charges,
• $k$ is Coulomb's constant, approximately $8.99 \times 10^9 , \text{N m}^2/\text{C}^2$,
• $q_1$ and $q_2$ are the charges.

We need to solve for $q_2$, the charge on the plastic bead.

3. Rearrange to Find $q_2$

Rearranging the formula gives us:

$$ |q_2| = \frac{F r^2}{k |q_1|} $$

4. Substitute the Values

Now, substitute the values into the equation:

$$ |q_2| = \frac{(8.0 \times 10^{-4} , \text{N})(0.029 , \text{m})^2}{(8.99 \times 10^9 , \text{N m}^2/\text{C}^2)(3.5 \times 10^{-9} , \text{C})} $$

5. Calculate the Charge $q_2$

Perform the calculations step-by-step:

• Calculate $F r^2$:

$$ (8.0 \times 10^{-4})(0.029)^2 = 8.0 \times 10^{-4} \times 0.000841 = 6.72 \times 10^{-7} $$

• Calculate $k |q_1|$:

$$ (8.99 \times 10^9)(3.5 \times 10^{-9}) = 31.4655 $$

• Now substitute into the equation for $|q_2|$:

$$ |q_2| = \frac{6.72 \times 10^{-7}}{31.4655} \approx 2.14 \times 10^{-8} , \text{C}$$

Convert to nanoCoulombs:

$$ |q_2| \approx 2.14 , \text{nC} $$

6. Determine the Sign of the Charge

Since the glass bead is positively charged, the plastic bead must be negatively charged to exert a repulsive force:

$$ q_2 \approx -2.14 , \text{nC} $$