Solveeit Logo

Question

Question: A test charge \(q\) is located \(1m\) from a much larger and stationary charge \(Q\). While at this ...

A test charge qq is located 1m1m from a much larger and stationary charge QQ. While at this location, the test charge, qq experiences a force of FF from the stationary charge QQ. The test charge is then moved to a new location 2m2m from QQ . What force will the test charge qq experience from the stationary charge QQ at the new location?
A.)(14)F\left( {\dfrac{1}{4}} \right)F
B.)(12)F\left( {\dfrac{1}{2}} \right)F
C.)FF
D.)2F2F

Explanation

Solution

Hint- You can start by describing electromagnetic force. Then you can describe Coulomb’s law and its equation. Use Coulomb’s law for both the initial and final conditions given in the problem, compare the two forces and find the solution.

Complete step-by-step answer:
Given in the problem,
ri=1m rf=2m  {r_i} = 1m \\\ {r_f} = 2m \\\

Let the initial and final force be Fi{F_i}andFf{F_f}

According to Coulomb’s law

F=kq1q2r2F = \dfrac{{k{q_1}{q_2}}}{{{r^2}}}
Here,

k=k = Proportionality constant
F=F = Electrostatic force
q1={q_1} = 1st charge
q2={q_2} = 2nd charge
r=r = Distance between the two charges

Applying Coulomb’s law for initial conditions

Fi=kQqri2{F_i} = \dfrac{{kQq}}{{r_i^2}}
F1=kQq(1)2{F_1} = \dfrac{{kQq}}{{{{(1)}^2}}}(Equation 1)

Applying Coulomb’s law for final conditions

Ff=kQqrf2{F_f} = \dfrac{{kQq}}{{r_f^2}}
Ff=kQq(2)2\Rightarrow {F_f} = \dfrac{{kQq}}{{{{\left( 2 \right)}^2}}}
Ff=kQq4\Rightarrow {F_f} = \dfrac{{kQq}}{4}
Ff=Fi4\Rightarrow {F_f} = \dfrac{{{F_i}}}{4}

Hence, option A is the correct choice

Additional Information:
The force that a charge experiences in the presence of another charge is known as electrostatic force. This force is also known as Coulomb force, after the name of its discoverer Charles-Augustin de Coulomb.

The charge experiences Coulomb’s force due to the formation of an electric field around the other charge. Imagine a fisherman entrapping a fish in its net and pulling it, this is somewhat how electric fields trap a charge.

Note: In the solution, we considered both QQ and to be positive, but, you can also consider charges negative. Changing the nature of charges will not affect the final answer, and it only means that the force can be attractive in case of opposite nature or it can be repulsive.