Point charges of 6.00 µc and −2.75 µc are placed 0.300 m apart. (assume the negative charge is located to the right of the positive charge. include the sign of the value in your answers.) (a) where can a third charge be placed so that the net force on it is zero?

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.

Put the value  into the formula

Hence, A third charged be placed at 0.63 m.