Question

Two positive charges ( 8.0 mC and 2.0 mC) are separated by 300 m. A third charge is placed at distance r from the 8.0 mC charge in such a way that the resultant electric force on the third charge due to the other two charges is zero. The distance r is

Answer 1

Answer:

$r=200m$

Explanation:

From the question we are told that:

Charges:

$Q_1=8.0mC$

$Q_2=2.0mC$

$Q_3=8.mC$

Distance $d=300m$

Generally the equation for Force is mathematically given by

$F=\frac{kq_1q_2}{r^2}$

Therefore

$F_{32}=F_{31}$

$\frac{q_2}{(300-r)^2}=\frac{q_1}{r^2}$

$\frac{2*10^{-3}}{(300-r)^2}=\frac{8*10^{-3}}{r^2}$

$r=2(300-r)$

$r=200m$