Charge q is one unit of distance away from the source charge s. Charge p is two times further away. The force exerted between S and q is ? The force exerted between S and p.

Charge q is 1 unit of distance away from the source charge S. Charge p is two times further away. The force exerted between S and q is the force exerted between S and p

a. 1/2

b. 2 times

c. 1/4

d. 4 times

4 times

Step-by-step explanation:

Given the following :

Distance between q and S (r) = 1 unit

Distance between P and S (r) = 2 × 1 unit = 2

According to coliumbs law:

F = Ke(q1q2) /r^2

Where F = electrostatic force, Ke = coloumbs constant

q1,q2 = charges, r = distance between two charges

For q and S :

Fsq = Ke(q1q2) / 1^2

Fsq = Ke(q1q2) - - - - - equation 1

For P and S :

Fsp = Ke(q1q2) / 2^2

Fsp = Ke(q1q2) / 4 - - - - - equation 2

Dibiding equation 1 by equation 2

Fsq/Fsp = Ke(q1q2) / Ke(q1q2)/4

Fsq / Fsp = Ke(q1q2) × 4 / Ke(q1q2)

Fsq / Fsp = 4 / 1

Using cross multiplication

Therefore, Fsq = 4 × Fsp

the answers are in the images, i received 100% on this assignment. hope this !

x = 23.78

step-by-step explanation: