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
Answer:
4 times
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