Question

A +5.75 uc and a -4.55 uc charge is placed 16.5 cm apart. Where can a third charge be placed so that it experiences no net force?​

Answer 1

For a charge to experience no net force,

F1 = F2

kQ1Q = kQ2Q

(d + x)² x²

where, Q1 = 5.75e-6C

Q2 = 4.55e-6C

Q = third charge

d = 16.5e-2 m

x = unknown distance

kQ1Qx² = kQ2Q(d+x)²

Q1x² = Q2(d²+2dx+x²)

Q1x² = Q2d² + 2Q2dx + Q2x²

Q1x² - 2Q2dx - Q2x² = Q2d²

x²(Q1 - Q2) = Q2d² + 2Q2dx

x² = Q2d² + 2Q2dx

Q1 - Q2

2Q2dx is a very small value thus it's approximated to zero.

x²=4.55e-6*(16.5e-2)²

5.75e-6 - 4.55e-6

square rooting both sides

x = 16.5e-2√4.55e-6

√(5.75e-6 - 4.55e-6)

x = 0.32m = 32cm

The 3rd charge must be placed at 32cm so it experiences no net charge.