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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?

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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
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