Answer to Question #138228 in Mechanics | Relativity for Manfe

Question

Three charged particles are located at the corners of an equilateral triangle with sides 2.00C, -4.00C, 7.00C subtended by an angle of 60.0 degree at one end . Calculate the total electric force on the 7.00C charge.

Answer 1

2C and 7C are of the same sign, so the charges will be repelled.

-4C and 7C are of the different signs, so the charges will be attracted.

$\overrightarrow{F}=\overrightarrow{F_1}+\overrightarrow{F_2}$

Let r is the length of the side of the triangle. So,

$F_1=|\overrightarrow{F_1}|=\frac{k\cdot|2C|\cdot|7C|}{r^2}=\frac{14kC^2}{r^2}$,

$F_2=|\overrightarrow{F_2}|=\frac{k\cdot|-4C|\cdot|7C|}{r^2}=\frac{28kC^2}{r^2}=2\cdot\frac{14kC^2}{r^2}=2F_1$, where $k=9\cdot10^9$.

According the law of cosines,

$F=|\overrightarrow{F}|=\sqrt{{F_1}^2+{F_2}^2-2{F_1}{F_1}cos60\degree}=\frac{14Ck^2}{r^2}\sqrt{1+4-2\cdot1\cdot2\cdot0.5}=\sqrt{3}\cdot\frac{14kC^2}{r^2}=\sqrt{3}F_1.$It can be seen, that $F^2+F_1^2=F_2^2,$ so $F$ is perpendicular to $F_1$ (or the side that contains charges 2C and 7C).