Question

A negative charge of 2.0 x 10^-4 C and a positive charge of 8.0 x 10^-4 C are separated by 0.30 m. What is the magnitude of the force between the charges? Is this force repulsive or attractive?

Answer 1

Answer:

$1.6*10^{4}$N

Attractive

Explanation:

$We're given:\\q_{1} = -2* 10^{-4} \\q_{2} = 8* 10^{-4}\\r= 0.30 m\\We know;\\E=\frac{kq_{1}q_{2} }{r^{2} } \\E=\frac{9*10^{9}*(-2* 10^{-4} ) (8* 10^{-4} ) }{(0.30)^{2} } \\E=-1.6*10^{4}$

It is attractive in nature since both charges are of oppositely charged(i.e. one is -ve and one is +ve)

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