Answer to Question #172726 in Electricity and Magnetism for Raphayael Ymran Mascariñas

Question

Three point charges are arranged along the y-axis in a vacuum. The topmost charge bears a charge of -4.0 µC, 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

Force due to topmost charge on middle charge

$F'=\frac{kqq'}{r^2}\\=\frac{9\times 10^{9}\times4\times10^{-6} \times3\times10^{-6} }{r^2}(j)\\=\frac{0.108}{r^2}(j)$

Similiarly

$F''=\frac{kqq''}{r'^2}\\=\frac{9\times 10^{9}\times3\times10^{-6} \times7\times10^{-6} }{r'^2}(-j) \\=\frac{0.189}{r'^2}(-j)$

Therefore total force is written as

$F=F'+F''\\=\frac{0.108}{r^2}(j)+\frac{0.189}{r'^2}(-j)$

So we can say total force is fully depend only on distance of charges.

Let $r'=r''$ =1m

Electric force become

$F=0.181N$ in -y axis direction

So magnitue is

F=0.181N

And direction is -Y axis.