Question

Charges of 7.2nC and 6.7nC are 32 cm apart. Find the equilibrium position for a -3.0nC charge?????

Answer 1

$F_1=k\frac{q_1q}{x^2}$

$F_2=k\frac{q_2q}{(0.32-x)^2}$

$F_1=F_2\to k\frac{q_1q}{x^2}=k\frac{q_2q}{(0.32-x)^2}\to\frac{q_1}{x^2}=\frac{q_2}{(0.32-x)^2}\to$

$q_1(0.32-x)^2=q_2x^2\to(q_1-q_2)x^2-0.64q_1x+0.1024q_1=0$

$0.5x^2-4.608x+0.73728=0\to x\approx0.163\ (m)=16.3\ (cm)$ . Answer