Three point charges q_1 = +2.0 \times 10^{-6}\text{ C}, q_2 = +2.0 \times 10^{-6}\text{ C}… — JAMB UTME

Three point charges $q_1 = +2.0 \times 10^{-6}\text{ C}$ , $q_2 = +2.0 \times 10^{-6}\text{ C}$ , and $q_3 = -4.0 \times 10^{-6}\text{ C}$ are placed along a straight line at positions $x = 0\text{ m}$ , $x = 0.30\text{ m}$ , and $x = 0.60\text{ m}$ , respectively. Taking Coulomb's constant $k = 9.0 \times 10^9\text{ N m}^2\text{ C}^{-2}$ , calculate the magnitude of the net electrostatic force acting on charge $q_2$ in newtons.

Answer: 1.2 N

Answer

The magnitude of the net electrostatic force acting on charge

q_2

1.20\text{ N}

The force exerted on

q_2

q_1

is repulsive (

0.40\text{ N}

directed to the right) because both charges are positive. The force exerted on

q_2

q_3

is attractive (

0.80\text{ N}

directed to the right) because

q_2

is positive and

q_3

is negative. Since both component forces act in the same direction, the total net force magnitude is

0.40\text{ N} + 0.80\text{ N} = 1.20\text{ N}

Step-by-Step Solution

Calculate the repulsive force exerted by

q_1

q_2

F_{12} = 0.40\text{ N}

pointing to the right

Like charges repel each other, so

q_1

pushes

q_2

away along the

+x

-axis.

Calculate the attractive force exerted by

q_3

q_2

F_{32} = 0.80\text{ N}

pointing to the right

Unlike charges attract each other, so

q_3

pulls

q_2

towards itself along the

+x

-axis.

Sum the component electrostatic forces acting on

q_2

F_{\text{net}} = 0.40\text{ N} + 0.80\text{ N} = 1.20\text{ N}

Because both forces act in the exact same direction along the line, their magnitudes add directly.

Key Concept

Coulomb's Law and Principle of Superposition for Electrostatic Forces

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