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Anna007 [38]
3 years ago
8

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 c

onstant 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.

Physics
1 answer:
igor_vitrenko [27]3 years ago
3 0

Answer:

0.64 m from the first charge

Explanation:

Force is given by

F_1=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F_1=\dfrac{k2\times 10^{-9}\times 3.7\times 10^{-9}}{x^2}

F_2=\dfrac{kq_2q_3}{r^2}\\\Rightarrow F_1=\dfrac{k4.5\times 10^{-9}\times 3.7\times 10^{-9}}{(1.6-x)^2}

These forces are equal

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

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

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