Question

Two charged particles are a distance of 1.82 m from each other. One of the particles has a charge of 8.22 nC, and the other has a charge of 4.02 nC.

(a)
What is the magnitude (in N) of the electric force that one particle exerts on the other?
N

Answer 1

The magnitude of the electric force between the two charged particles is 29.94 N.

The magnitude of the electric force between two charged particles can be calculated using Coulomb's Law.

Coulomb's Law states that the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it can be represented as:

F =
`(kq_(1)q_(2))/(r^(2))`

where F is the electric force,
`q_(1)` and
`q_(2)` are the charges of the particles, r is the distance between them, and k is the electrostatic constant (k = 8.99 *
`10^(9)` Nm²/C²).

Using the given values:

F =
`(8.22*10^(-9) * 4.02 * 10^(-9))/(1.82^(2))`

F = 29.94 N

Therefore, the magnitude of the electric force that one particle exerts on the other is 29.94 N.

Answer 2

The magnitude of the electric force on one of the particles is 8.98 x 10⁻⁸ N.

How to calculate the magnitude of electric force on the particles?

The magnitude of the electric force on one of the particles is calculated by applying Coulomb's law of electrostatic force;

F = kq₁q₂ / r²

where;

• q₁ and q₂ are the magnitude of first and second charge respectively
• r is the distance between the charges
• k is Coulomb's constant

The magnitude of the electric force on one particle is calculated as;

F = (9 x 10⁹ x 8.22 x 10⁻⁹ x 4.02 x 10⁻⁹ ) / ( 1.82 m)²

F = 8.98 x 10⁻⁸ N

Thus, the magnitude of the electric force on one of the particles is 8.98 x 10⁻⁸ N.