Question

One charge +5.00 µC charged particle is at x = 0.00 m, and the second charge(+7.00 µC) farther along the x-axis. The net force from the charges is zero at x=45.8cm. Where is the +7.00 µC located on the x axis?

Answer 1

Electrostatic force is given by:

F = k*q1*q2/r^2

Net force at x = 45.8 cm will be on third charge q will be

Fnet = F1 + F2

Since both charges are positive and net force is zero, So this means that third charge q is between both given charges at x = 45.8 cm. Suppose charge q2 = +7.00 uC is at x = d cm

Fnet = F1 + F2 = 0

F1 = -|F2|

k*q1*q/r1^2 = k*q2*q/r2^2

q1/r1^2 = q2/r2^2

q1 = 5 uC & q2 = 7 uC

r1 = 45.8 cm = 0.458 m

r2 = d - 0.458 m

So,

5/0.458^2 = 7/(d - 0.458)^2

d - 0.458 = 7*0.458^2/5

d = 0.752 m = 75.2 cm

So charge 7.00 uC is located at = x = +75.2 cm

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