Two electrostatic point charges of +60.0 uC and +50.c uC exert a repulsive force on each other of 175 N. What is the distance between the two charges?

Answer 1

Here's what it looks like when you plug in the numbers for F = kqQ / r² :

k = 9 x 10⁹ N^{m^2} / C²

Q = 60 µC = 6 x 10^-5 C

Q' = 50 µ√C = 5 x 10^-5 C

175 = 9 x 10⁹ (6 x 10^-5)(5 x 10^-5 C) / r²

175 = (270)(10^-1) / r²

r² = 2.7 x 10 / 1.75 x 10² = 1.54 x 10^-1 m²

r = √1.54x10^-1 ≈ 12 cm

Answer 2

Consider the force between two point charges:

F_q = k·q₁·q₂ / r²,

where we know the values for all of the variables except for r, and so we solve for it:

r = √(k·q₁·q₂ / F_q).

Answer 3

F = kqQ / r² ⇒

r = √(kqQ / F)

q = 50 x 10^-6 C

Q = 60 x 10^-6 C

F = 175 N

k = 9 x 10⁹ Nm²/C²

Plug in the numbers and get r in meters.