Question

Three point charges \( +q, +2q, \) and \( +4q \) are placed along a straight line such that the charge \( +2q \) lies equidistant from the other two charges. The ratio of the net electrostatic force on charges \( +q \) and \( +4q \) is:

• A. \(1 : 1\)
• B. \(1 : 2\)
• C. \(1 : 4\)
• D. \(1 : 3\)

Hint:

In electrostatics, when calculating the net force on charges, remember to consider both the magnitudes and directions of the forces due to each charge and sum them vectorially.

Answer 1

The Correct Option is D

Let the charges \( +q \) and \( +4q \) be at positions \( x = -a \) and \( x = +a \) respectively, with \( +2q \) at the origin. The forces due to each charge on the other are given by Coulomb's law: \[ F = k \frac{|q_1q_2|}{r^2} \] where \( k \) is Coulomb's constant. 
Force on \( +q \): The force on \( +q \) due to \( +2q \) is: \[ F_{q \to 2q} = k \frac{|q \cdot 2q|}{a^2} = k \frac{2q^2}{a^2} \] Similarly, the force on \( +q \) due to \( +4q \) is: \[ F_{q \to 4q} = k \frac{|q \cdot 4q|}{(2a)^2} = k \frac{4q^2}{4a^2} = k \frac{q^2}{a^2} \] Thus, the net force on \( +q \) is: \[ F_{{net}, q} = F_{q \to 2q} + F_{q \to 4q} = k \frac{2q^2}{a^2} + k \frac{q^2}{a^2} = k \frac{3q^2}{a^2} \] 
Force on \( +4q \): The force on \( +4q \) due to \( +2q \) is: \[ F_{4q \to 2q} = k \frac{|4q \cdot 2q|}{a^2} = k \frac{8q^2}{a^2} \] Similarly, the force on \( +4q \) due to \( +q \) is: \[ F_{4q \to q} = k \frac{|4q \cdot q|}{(2a)^2} = k \frac{4q^2}{4a^2} = k \frac{q^2}{a^2} \] Thus, the net force on \( +4q \) is: \[ F_{{net}, 4q} = F_{4q \to 2q} + F_{4q \to q} = k \frac{8q^2}{a^2} + k \frac{q^2}{a^2} = k \frac{9q^2}{a^2} \] 
Ratio of Forces: The ratio of the net forces on \( +q \) and \( +4q \) is: \[ {Ratio} = \frac{F_{{net}, q}}{F_{{net}, 4q}} = \frac{k \frac{3q^2}{a^2}}{k \frac{9q^2}{a^2}} = \frac{3}{9} = \frac{1}{3} \] Thus, the ratio of the net electrostatic forces is \( 1 : 3 \).