Question

A charge q1 of +8.00 × 10−9 C and a charge q2 of +4.00 × 10−9 C are separated by a distance

of 60.0 cm. What is the equilibrium position for a third charge of

–25.0 × 10−9 C?

Answer 1

Answer

Force at equilibrium point is equal so

$\frac{kq_1q_3}{r^2}=\frac{kq_2q_3}{r'^2}$

$\frac{8\times10^{-9}(-25)\times10^{-9}}{x^2}=\frac{ 4\times10^{-9} (-25) \times10^{-9} }{(0.6-x)^2}$

$2(0.36+x^2-1.2x) =x^2$

Or we can write this as

$\sqrt{2}(0.6-x) =x$

$0.84-1.4x=x$

$2.4x=0.84$

$x=0.4m=40cm$

So the equilibrium position for a third charge of

–25.0 × 10^{−9} C

At 40cm from +8.00 × 10^{−9} C.