Answer to Question #295538 in Electricity and Magnetism for eli

Question

three point charges, which are equally separated by a distance 20 cm, are arranged along the y-axis in a vacuum. The topmost charge bears a charge of – 4.0μC, the middle charge has a charge of +3.0 μC, and the bottom one carries a 7.0 μC charge. What is the magnitude and direction of the net electrostatic force that the middle charge experiences?

Answer 1

Answer

Charge

Q=-4$\mu C$

$Q'=3\mu C$

$Q''=7\mu C$

So all are seperated by d=0.2m

Net electrostatic force that the middle charge

"F=\\frac{KQQ'}{d^2}+\\frac{KQ"Q'}{d^2}"

So putting all values

$F=\frac{9\times10^{9}\times4\times10^{-6} 3\times10^{-6} }{(0.2)^2}+\frac{9\times10^{9}\times 3\times10^{-6} \times7\times10^{-6} }{0.2\times0.2}$

So force is

$F=7.425N$