Question

A glass bead charged to 3.7nc exerts an 7.0×10−4n repulsive electric force on a plastic bead 2.9cm away. what is the charge on the plastic bead?

Answer 1

The electrostatic force between the glass bead and the plastic bead is given by:
$F=k \frac{q_1 q_2}{r^2}$
where
k is the Coulomb's constant
q1 is the charge of the glass bead
q2 is the charge of the plastic bead
r is the separation between the two beads

In this problem, $r=2.9 cm=0.029 m$ and $q_1 = 3.7 nC=3.7 \cdot 10^{-9} C$, and since we know the force, $F=7.0 \cdot 10^{-4}N$, we can rearrange the equation to find q2, the charge of the plastic bead:
$q_2 = \frac{Fr^2}{kq_1}= \frac{(7.0 \cdot 10^{-4}N)(0.029 m)^2}{(9.0 \cdot 10^9 Nm^2C^{-2})(3.7\cdot 10^{-9}C)}=1.77 \cdot 10^{-8}C$

Answer 2

The charge on the plastic bead is 17.68 nC

Data obtained from the question

• Charge on glass bead (q₁) = 3.7 nC = 3.7×10¯⁹ C
• Force (F) = 7×10¯⁴ N
• Distance apart (r) = 2.9 cm = 2.9 / 100 = 0.029 m
• Electrical constant (K) = 9×10⁹ Nm²/C²
• Charge on plastic bead (q₂) =?

How to determine the charge on the plastic bead

F = Kq₁q₂ / r²

7×10¯⁴ = (9×10⁹ × 3.7×10¯⁹ × q₂) / (0.029)²

Cross multiply

9×10⁹ × 3.7×10¯⁹ × q₂ = 7×10¯⁴ × (0.029)²

Divide both side by 9×10⁹ × 3.7×10¯⁹

q₂ = [7×10¯⁴ × (0.029)²] / (9×10⁹ × 3.7×10¯⁹)

q₂ = 17.68×10¯⁹ C

q₂ = 17.68 nC

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