I need help solving this question.
"Two charges are 10 meters apart. The charge on the positive particle is 1.0 C and the negative charge is 10 Coulombs. Calculate the magnitude of the electrostatic force of attraction between these two charges."
Use Coulomb's Law
F = k Q1 Q2/R^2
k = 8.99*10^9 N/m^2 C^2
Q1 = 1 C
Q2 = 10 C
R = 10 m
Your answer will be in Newtons. Never mind the minus sign on Q2. The force F will be an attraction since the charges are opposite.
To calculate the magnitude of the electrostatic force of attraction between two charges, you can use Coulomb's law. Coulomb's law states that the magnitude of the electrostatic force between two charged objects is directly proportional to the product of their 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 electrostatic force of attraction between the charges,
k is the electrostatic constant (which is approximately equal to 9 x 10^9 N*m^2/C^2),
q1 and q2 are the magnitudes of the two charges, and
r is the distance between the charges.
In this case, q1 is 1.0 C (charge on the positive particle) and q2 is -10 C (charge on the negative particle). The distance between the charges, r, is given as 10 meters.
Plugging in the values into the formula, we get:
F = (9 x 10^9 N*m^2/C^2 * |1.0 C * -10 C|) / (10 m)^2
Simplifying this expression gives:
F = 9 x 10^9 N*m^2/C^2 * 10 C^2 / (10 m)^2
Now we can calculate the value using a calculator:
F = 9 x 10^9 N*m^2/C^2 * 10 C^2 / (10 m)^2 = 9 x 10^9 N
Therefore, the magnitude of the electrostatic force of attraction between the two charges is 9 x 10^9 Newtons.