Question

Two charged particles attract each other with a force of magnitude F acting on each. If the charge of one is doubled and the distance separating the particles is also doubled, the force acting on each of the two particles has magnitude

(a) F/2

(b) F/4

(c) F

(d) 2F

(e) 4F

(f) None of the above.

 

Answer 1

According to coulomb’s law, the force is,

F = 1/4πε0 ∙ q2/r2

 

Here, ε0 is the permittivity of free space, q is the charge of the particle, and r is the distance between the charges.

 

If the charge of one is doubled and the distance of separating the particles is also doubled, the new charge and distances are as follows:

q2 = 2q1

r12 = 2r

 

The new force is,

F\' = 1/4πε0 ∙ q1q2/r212

 

Here, q1 is the charge of first particle, q2 is the charge of second particle, and r12 is the distance between the charges.

 

Substitute q for q1, 2r for r12 and 2q for q2 in the above equation.

F\' = 1/4πε0 ∙ q(2q)/(2r)2

    = 1/4πε0 2q2/4r2

   = ½(1/4πε0 ∙ q2/r2)

 

Substitute F for 1/4πε0 ∙ q2/r2 in the above equation.

F\' F/2

 

Hence, the options (b), (c), (d), (e), and (f) are not correct.

Hence, the correct option is (a).

The correct option is (a).