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.

Answer 1

The magnitude of the force between two equal charges each of 1 C each separated by a distance of 1 km is 9 x 10^9 N.

Step-by-step explanation:

To calculate the magnitude of the force between two equal charges each of 1 C each separated by a distance of 1 km in air, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. The equation for Coulomb's Law is:

F = k * (q1 * q2) / r^2

Where F is the magnitude of the force, k is Coulomb's constant (approximately 9 x 10^9 N * m^2 / C^2), q1 and q2 are the charges, and r is the distance between the charges.

Plugging in the values given, we get:

F = (9 x 10^9 N * m^2 / C^2) * (1 C * 1 C) / (1 km)^2

Converting the distance to meters:

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

Simplifying the equation, we get the magnitude of the force between the charges:

F = 9 x 10^9 N

Answer 2

9000 N

Step-by-step explanation:

Charge, q1 = q2 = 1 C

Distance, r = 1 km = 1000 m

According to the coulomb's law, the force between the two charged particles is given by

`F =(1)/(4\pi \varepsilon _(0))\frac{q_(1)q_(2)}}{r^(2)}`

By substituting the values

`F =(9* 10^(9)* 1* 1)/(1000* 1000)`

F = 9000 N

Thus, the force of repulsion between the charges is 9000 N.