Two point charges, +3.56 μC and -1.93 μC attract each other with a force of magnitude 0.292 N. What is their separation?

Given,

Charge (Q) = +3.56*10-6C

Charge (q) = -1.93*10-6C

K (constant) = 8.99*109Nm²/c²

Force (F) = 0.292N

Distance between two point charges (r) =?

We know that:-

From Coulomb’s law:-

F = KQq/r²

0.292 = 8.99*109*(3.56*10-6)*(-1.93*10-6)/r²

0.292*r² = +8.99*109*3.56*1.93*10-12

R2 = +8.99*109*3.56*1.93*10-12/0.292

R2 = +211.5359*10-3

R = √211.5359*10-3 = 0.459m

R = 0.459m

Distance between two point charges = 0.459m

According to Coulomb,
F = kQq / r²
and a negative force indicates attraction, so
-0.292N = 8.99e9N·m²/C² * 3.56e-6C * -1.93e-6C / r²
solves to
r = 0.460 m ◄

Hope this helps!

F=kQ1Q1/r

K=9.0*10^9
r = distance

Get R by itself

F=kQ1Q1/r

FR=KQ1Q2

R=KQ1Q2/F

R=(9.0*10^9)(3.56*10^-6)(-1.93*10^-6) / (0.292)

R = 0.21m