Question

Two positive charges of 6.0 x 10^-6 C are separated by 0.50 m. Calculate the force of electrical attraction

Answer 1

Answer:

Force between two charges is given as 1.296 N

Explanation:

As we know that the two charges will attract or repel each other when they are placed near to each other

Here we know that electrostatic force is given as

$F = \frac{kq_1 q_2}{r^2}$

here we have

$q_1 = q_2 = 6 \times 10^{-6} C$

distance between two charges is given as

r = 0.50 m

so we have

$F = \frac{(9\times 10^9)(6 \times 10^{-6})^2}{0.50^2}$

$F = 1.296 N$