Question

Two charges of 5 x 10-6 C are a certain distance apart. These two positive charges experience 3.6 N of repelling electrical force between them. How far apart are these charges?

Answer 1

Answer:

r = 0.2499 m

Explanation:

Use Coulomb's Law: $F_{e} = kq_{1}q_{2} / r^{2}$ where Fe is the electric force, q1 and q2 are the 2 charges respectively, k is a constant (8.99*10^9 N • m^2 / C^2) and r is the distance.

$3.6 N = (8.99*10^{9})(5*10^{-6})(5*10^{-6}) / r^{2}$

r = 0.2499 m

Answer 2

The distance that should be apart from these charges should be considered when the r = 0.2499 m.

Calculation of the distance apart:

Since

Two charges of 5 x 10-6 C are a certain distance apart. These two positive charges experience 3.6 N of repelling electrical force between them

So

here we use Columb law

Electric force = k * two charges * two charges)r^2

3.6 N = (8.99*10^9) * ( 5 * 10^-6) * (5*10^-6)r^2

r = 0.2499 m.

Hence, The distance that should be apart from these charges should be considered when the r = 0.2499 m.

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