Question

At what separation will two charges, each of magnitude 6.0fYC,exert a force of 1.4N on each other?

a.0.48m

b.2.0m

c.5.1*10^6m

d.0.23m

e.40m

Answer 1

I assume that "fYC" is just a writing mistake and the two charges have magnitude $q=6 \mu C=6 \cdot 10^{-6}C$.
The electrostatic force between the two charges is
$F=k_e \frac{q^2}{d^2}$
where $k_e = 8.99 \cdot 10^9 Nm^2C^{-2}$ is the Coulomb's constant, q is the magnitude of the two charges, and d is the separation between them.

We know the value of the force, F=0.14 N, so re-arranging the formula and using these data we can solve to find the value of d:
$d= \sqrt{ \frac{k_e q^2}{F} } =0.48 m$

So, the separation between the two charges is 0.48 m.

Answer 2

This question involves the concept of Colomb's Law and electrostatic force.

The separation between charges will be "a. 0.48 m".

COLOMB'S LAW

According to Colomb's Law, every charge exerts an electrostatic force on the other charge, which is directly proportional to the product of the magnitudes of both the charges and inversely proportional to the square of the distance between them.

$F = \frac{kq_1q_2}{r^2}$

where,

• F = electrostatic force = 1.4 N

• k = Colomb's constant = 9 x 10⁹ N.m²/C²

• q₁ = magnitude of first charge = 6 μC = 6 x 10⁻⁶ C

• q₂ = magnitude of second charge = 6 μC = 6 x 10⁻⁶ C

• r = distance between charges = ?

Therefore,

$1.4 N = \frac{(9\ x\ 10^9\ N.m^2/C^2)(6\ x\ 10^{-6}\ C)(6\ x\ 10^{-6}\ C)}{r^2}\\\\r=\sqrt{\frac{(9\ x\ 10^9\ N.m^2/C^2)(6\ x\ 10^{-6}\ C)(6\ x\ 10^{-6}\ C)}{1.4\ N}}$

r = 0.48 m

Learn more about Colomb's Law here:

brainly.com/question/9774180