Respuesta :
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[tex]F_1 and F_2[/tex]
where [tex]F_1[/tex]=force exerted by first positive charge on negative charge
[tex]F_2[/tex]=force exerted by second positive charge on negative charge
[tex]F_1=F_2=\frac{kq_1q_2}{r^2}[/tex]
[tex]=\frac{9\times 10^9\times 4.5^2\times 10^{-12}}{(10^{-2})^2}[/tex]
[tex]=1822.5 N[/tex]
Two are at an angle of 60 (as it is placed at a corner of equilateral triangle)
[tex]F_{net}=\sqrt{1882.5^2+1882.5^2+2\times 1882.5\times 1882.5\times \cos 60}[/tex]
[tex]F_{net}=\sqrt{3\times 1882.5^2}[/tex]
[tex]F_{net}=\sqrt{3}1882.5[/tex]
Force Experienced by a positive charge will be of attractive and repulsive in nature
Two Forces are at an angle of 120 therefore
[tex]F_{net}=\sqrt{1882.5^2+1882.5^2+2\times 1882.5\times 1882.5\times \cos 120}[/tex]
[tex]F_{net}=1882.5 N[/tex]
