Question

An attractive force of 7.2 N occurs between two charges that are 0.20 m apart. If one charge is 4.0 μC, what is the other charge?

Answer 1

F = ko *Q_1 * Q_2 / d^2

Q_2 = F*d^2 / [ko*Q_1] = 7.2 N * (0.2m)^2 / [9.0 x 10^9 N/m^2C^2 * 4.0 x 10^-6C] = 8x10^-6C = 8 microC

Answer 2

The other charge of the particle given the data from the question is 8 μC

Data obtained from the question

• Force (F) = 7.2 N
• Distance apart (r) = 0.2 m
• Electrical constant (K) = 9×10⁹ Nm²/C²
• Charge on 1st particle (q₁) = 4 μC = 4×10¯⁶ C
• Charge on 2nd particle (q₂) = ?

How to determine the 2nd charge

The charge on the other particle can be obtained by using the Coulomb's law equation as shown below:

F = Kq₁q₂ / r²

7.2 = (9×10⁹ × 4×10¯⁶ × q₂) / 0.2²

7.2 = (9×10⁹ × 4×10¯⁶ × q₂) / 0.04

Cross multiply

9×10⁹ × 4 × q₂ = 7.2 ×  0.04

Divide both side by 9×10⁹ × 4×10¯⁶

q₂ = (7.2 ×  0.04) / (9×10⁹ × 4×10¯⁶)

q₂ = 8×10¯⁶ C

q₂ = 8 μC

Learn more about Coulomb's law:

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