213k views
3 votes
The force of attraction between a -35.0 μC and +101 μC charge is 4.00 N. What is the separation 15) between these two charges?

1 Answer

2 votes

Answer: the separation between the two charges is approximately 111.70 meters.

Step-by-step explanation:

To find separation between the two charges, we can use Coulomb's Law,

Coulomb's Law equation:

F = k * |q1 * q2| / r^2

Given:

Charge q1 = -35.0 μC (microCoulombs)

Charge q2 = +101 μC (microCoulombs)

Force F = 4.00 N

k is the electrostatic constant (k = 8.99 x 10^9 N m^2/C^2)

Rearranging the equation to solve for separation distance (r):

r = sqrt(k * |q1 * q2| / F)

Plugging in the values:

r = sqrt((8.99 x 10^9 N m^2/C^2) * |(-35.0 μC) * (+101 μC)| / 4.00 N)

r = sqrt((8.99 x 10^9 * 35.0 * 101) / 4.00) m

r ≈ 111.70 meters

Therefore, the separation between the two charges is approximately 111.70 meters.

User Hikmet
by
8.0k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.