A point charge of +6.0 \times 10^{-6}\text{ C} experiences an attractive electrostatic for… — JAMB UTME

A point charge of $+6.0 \times 10^{-6}\text{ C}$ experiences an attractive electrostatic force of $0.54\text{ N}$ when placed at a distance of $1.0\text{ m}$ from a second point charge in a vacuum. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}$ , what is the magnitude of the second charge in microcoulombs ( $\mu\text{C}$ )?

Answer: 10 µC

Answer

The magnitude of the second charge is 10 µC.

Using Coulomb's law

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

, substituting

F = 0.54\text{ N}

q_1 = 6.0 \times 10^{-6}\text{ C}

r = 1.0\text{ m}

, and

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

yields

|q_2| = 1.0 \times 10^{-5}\text{ C}

, which equals

10\text{ }\mu\text{C}

Step-by-Step Solution

State Coulomb's Law formula

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

Coulomb's Law describes the electrostatic force between two point charges.

Substitute given parameters into the equation

0.54 = \frac{9.0 \times 10^9 \times 6.0 \times 10^{-6} \times |q_2|}{1.0^2}

Knowns:

F = 0.54\text{ N}

q_1 = 6.0 \times 10^{-6}\text{ C}

r = 1.0\text{ m}

k = 9.0 \times 10^9\text{ N}\cdot\text{m}^2\text{C}^{-2}

Solve for the unknown charge magnitude

q_2

|q_2| = 1.0 \times 10^{-5}\text{ C}

Rearranging yields

|q_2| = \frac{0.54}{5.4 \times 10^4} = 1.0 \times 10^{-5}\text{ C}

Convert the value from Coulombs to microcoulombs

10\text{ }\mu\text{C}

1\text{ }\mu\text{C} = 10^{-6}\text{ C}

, so

1.0 \times 10^{-5}\text{ C} = 10\text{ }\mu\text{C}

Key Concept

Coulomb's Law

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