Question

If two equal charges each of 1 C each are separated in air by a distance of 1 km, what is the magnitude of the force acting between them? You will see that even at a distance as large as 1 km, the repulsive force is substantial because 1 C is a very significant amount of charge.

a) ( 8.99 X 10⁹ N )
b) ( 2.25 X 10⁸ N )
c) ( 1.80 X 10⁹ N )
d) ( 4.50 X 10⁸ N )

Answer 1

Using Coulomb's Law, the electrostatic force between two charges of 1 C each separated by 1 km in air is calculated to be 8.99 × 10^9 N.

Step-by-step explanation:

When two equal charges, in this case, each with a magnitude of 1 Coulomb (C), are separated by a distance of 1 kilometer (km) in air, we can calculate the force between them using Coulomb's Law. Coulomb's Law states that the force (F) between two charges (q1 and q2) is directly proportional to the product of the charges and inversely proportional to the square of the distance (r) between them. The formula is given by:

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

Where:

• k is Coulomb's constant (8.9875 × 10^9 N m^2/C^2)
• q1 and q2 are the magnitudes of the charges, both are 1 C in this scenario
• r is the separation distance, which is 1000 m (1 km converted to meters)

Substituting the given values, we get:

F = (8.9875 × 10^9 N m^2/C^2) * (1 C * 1 C) / (1000 m)^2

Which simplifies to:

F = 8.9875 × 10^9 N

Therefore, the correct answer is (a) (8.99 × 10^9 N).