113k views
4 votes
A charge of +2.00 x 10^-9 C is placed at the origin, and another charge of +4.50 x 10^-9 C is placed at x = 1.6 m. The Coulomb constant is 8.98755 x 10^9 N m2/C2. Find the point (coordinate) between these two charges where a charge of +3.70 x 10^-9 C should be placed so that the net electric force on it is zero.

1 Answer

0 votes

Answer:

0.64 m from the first charge

Step-by-step explanation:

Force is given by


F_1=(kq_1q_2)/(r^2)\\\Rightarrow F_1=(k2* 10^(-9)* 3.7* 10^(-9))/(x^2)


F_2=(kq_2q_3)/(r^2)\\\Rightarrow F_1=(k4.5* 10^(-9)* 3.7* 10^(-9))/((1.6-x)^2)

These forces are equal


(k2* 10^(-9)* 3.7* 10^(-9))/(x^2)=(k4.5* 10^(-9)* 3.7* 10^(-9))/((1.6-x)^2)\\\Rightarrow (2)/(x^2)=(4.5)/((1.6-x)^2)\\\Rightarrow (2)/(4.5)=(x^2)/((1.6-x)^2)\\\Rightarrow (4.5)/(2)=((1.6-x)^2)/(x^2)\\\Rightarrow \sqrt{(4.5)/(2)}=(1.6-x)/(x)\\\Rightarrow 1.5=(1.6-x)/(x)\\\Rightarrow 1.5x+x=1.6\\\Rightarrow x=(1.6)/(2.5)\\\Rightarrow x=0.64\ m

The distance that charge should be placed is 0.64 m from the first charge

A charge of +2.00 x 10^-9 C is placed at the origin, and another charge of +4.50 x-example-1
User Desha
by
2.7k points