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 Coulomb's Law
   Coulomb's Law states that the electric force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by:
   $$ F = k \frac{|q_1 q_2|}{r^2} $$
   where $k$ is Coulomb's constant, approximately $8.99 \times 10^9 , \text{N m}^2/\text{C}^2$.

2. Input Given Values
   We know:

• $q_1 = 3.5 , \text{nC} = 3.5 \times 10^{-9} , \text{C}$
• $F = 8.0 \times 10^{-4} , \text{N}$
• $r = 2.9 , \text{cm} = 0.029 , \text{m}$
  We need to find $q_2$, the charge on the plastic bead.

3. Rearrange Coulomb's Law
   To find $q_2$, we rearrange the equation:
   $$ |q_2| = \frac{F r^2}{k |q_1|} $$

4. Substitute Known Values
   Now plug in the known values:
   $$ |q_2| = \frac{(8.0 \times 10^{-4} , \text{N})(0.029 , \text{m})^2}{(8.99 \times 10^9 , \text{N m}^2/\text{C}^2)(3.5 \times 10^{-9} , \text{C})} $$

5. Calculate the Expression
   Calculate the numerator and the denominator:

• Numerator:
  $$ 8.0 \times 10^{-4} \times (0.029)^2 = 8.0 \times 10^{-4} \times 0.000841 = 6.72 \times 10^{-7} , \text{N m}^2 $$

• Denominator:
  $$ 8.99 \times 10^9 \times 3.5 \times 10^{-9} = 31.465 \approx 31.5 , \text{N m}^2/\text{C} $$

Now divide the two.

6. Final Calculation for $|q_2|$
   $$ |q_2| = \frac{6.72 \times 10^{-7}}{31.5} \approx 2.13 \times 10^{-8} \text{C} \approx 2.1 \text{nC} $$

Since the original charge of the glass bead is positive, the plastic bead must also be positively charged to exert a repulsive force.