Answer :
Answer:
A third charged be placed at 0.63 m.
Explanation:
Given that,
Point charges
q₁ = 6.00 μC
q₂=-2.75 μC
Distance = 0.300 m
(a). We need to calculate the distance of third charge
Using formula of net force
The net force will be zero at the point where the electric field intensity will be zero.
[tex]\dfrac{kq_{2}q_{3}}{d^2} =\dfrac{kq_{1}q_{3}}{(0.300+d)^2}[/tex]
Put the value into the formula
[tex]\dfrac{(0.300+d)^2}{d^2}=\dfrac{kq_{1}q_{3}}{kq_{2}q_{3}}[/tex]
[tex]\dfrac{(0.300+d)^2}{d^2}=|\dfrac{6.00\times10^{-6}}{-2.75\times10^{-6}}|[/tex]
[tex]\dfrac{(0.300+d)}{d}=\sqrt{2.18}[/tex]
[tex]\dfrac{0.300+d}{d}=1.47[/tex]
[tex]0.300=1.47d-d[/tex]
[tex]d=\dfrac{0.300}{0.47}[/tex]
[tex]d=0.63\ m[/tex]
Hence, A third charged be placed at 0.63 m.