Question

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equilateral triangle, as the drawing shows. The magnitude of each of the charges is 4.5 μC, and the lengths of the sides of the triangle are 1.0 cm. Calculate the magnitude of the net force that each charge experiences.

Answer 1

Answer:

Explanation:

Given

magnitude of charge=4.5\nu C

length of side of triangle=1 cm

There are two Positive charge and one negative charge

So net Force on negative charge is

sum of$F_1 and F_2$

where $F_1$=force exerted by first positive charge on negative charge

$F_2$=force exerted by second positive charge on negative charge

$F_1=F_2=\frac{kq_1q_2}{r^2}$

$=\frac{9\times 10^9\times 4.5^2\times 10^{-12}}{(10^{-2})^2}$

$=1822.5 N$

Two are at an angle of 60 (as it is placed at a corner of equilateral triangle)

$F_{net}=\sqrt{1882.5^2+1882.5^2+2\times 1882.5\times 1882.5\times \cos 60}$

$F_{net}=\sqrt{3\times 1882.5^2}$

$F_{net}=\sqrt{3}1882.5$

Force Experienced by a positive charge will be of attractive and repulsive in nature

Two Forces are at an angle of 120 therefore

$F_{net}=\sqrt{1882.5^2+1882.5^2+2\times 1882.5\times 1882.5\times \cos 120}$

$F_{net}=1882.5 N$