Question

A thundercloud has an electric charge of 43.2 C near the top of the cloud and -38.7 C near the bottom of the cloud. The magnitude of the electric force between these two charges is 3.95 x 106 N. What is the average separation between these charges? (kc = 8.99 ×109 N • m2 / C2)

Answer 1

1950.65m

Step-by-step explanation:

Coulomb's law states that the force F of attraction between two charges q1 and q2 is directly proportional to the product of the charges and inversely proportional to the square of the distance d between them. Mathematically;

|F| = k |q1||q2| /r²

Given |q1| = |43.2| C = 43.2C

|q2| = |-38.7|C = 38.7C

F = 3.95 x 106 N

d = ?

k is the coloumbs constant = 8.99 ×109 Nm2 / C2

Substituting this values in the formula to get distance d we have;

3.95x10^6= 8.99×10^9×43.2×38.7/d²

3.95x10^6d² = 15,029,841,600,000

d² =15,029,841,600,000/3,950,000

d² = 3,805,023.189

d = √3,805,023.189

d = 1950.65m

Average separation between the charges is 1950.65m

Answer 2

1950.65 m

Step-by-step explanation:

Parameters given:

First charge, Q1 = 43.2 C

Second charge, Q2 = -38.7 C

Magnitude of force between them, F = 3.95 * 10^6 N

The magnitude of the electric force between two electric charges at a distance r from one another is given as:

|F| = |k * Q1 * Q2|/r²

3.95 * 10^6 = |8.99 * 10^9 * 43.2 * -38.7| / r²

Making r² the subject of the formula and solving:

r² = 3805025.189

r = 1950.65 m