What is the electric force of attraction between charges of 3 C and -4 C separated by a distance of 2 m?
Coulomb’s Law
F = k •q1•q2/r²,
where
k =9•10^9 N•m^2/C^2,
q1 = 3 C,
q2 = |q2| = 4 C,
r =2 m
To calculate the electric force of attraction between two charges, you can use Coulomb's law. Coulomb's law states that the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's law is:
F = k * (|q1| * |q2|) / r^2
Where:
- F is the magnitude of the electric force between the charges
- k is the electrostatic constant, which has a value of approximately 9 x 10^9 Nm^2/C^2
- q1 and q2 are the magnitudes of the charges
- r is the distance between the charges
In this case, the charges are q1 = 3 C and q2 = -4 C, and the distance is r = 2 m.
Plugging the values into the formula:
F = (9 x 10^9 Nm^2/C^2) * (|3 C| * |-4 C|) / (2 m)^2
F = (9 x 10^9 Nm^2/C^2) * (12 C^2) / 4 m^2
F = (9 x 10^9 Nm^2/C^2) * 12 / 4
F = (9 x 10^9 Nm^2/C^2) * 3
F = 27 x 10^9 Nm^2/C^2
Therefore, the electric force of attraction between charges of 3 C and -4 C separated by a distance of 2 m is 27 x 10^9 N.