Using Coulomb's law to solve: Charge A is +2.0x10^-6C and charge B is +3.0x10^-6C. The two changes are 3 m apart. What is the force between them?Is the force attractive or repulsive?
Okay so here's what I have so far
(2.0x10^-6)(3.0x10^-6)/(3)^2
It is confusing me with the exponents. I know this is repulsive though already from the charges.
To solve this problem using Coulomb's law, you need to plug the given values into the formula and calculate the force between the charges. Coulomb's law states that the force between two charges is given by:
F = (k * q1 * q2) / r^2
Where F is the force, k is Coulomb's constant (approximately 9.0 x 10^9 Nm^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
In your case, q1 = +2.0x10^-6 C, q2 = +3.0x10^-6 C, and r = 3 m.
Let's calculate the force:
F = (9.0 x 10^9 Nm^2/C^2) * (2.0x10^-6 C) * (3.0x10^-6 C) / (3 m)^2
To perform calculations with exponents, you can use the rules of exponents:
Product rule: a^x * a^y = a^(x+y)
Quotient rule: a^x / a^y = a^(x-y)
Using these rules, we can simplify the above expression:
F = (9.0 x 2.0 x 3.0) x (10^9 x 10^-6 x 10^-6) / (3^2)
F = 54 x (10^(-6+(-6+9))) / 9
F = 54 x 10^(9-6-6) / 9
F = 54 x 10^(-3) / 9
F = 6 x 10^(-3) N
So, the force between the charges is 6 x 10^(-3) Newtons.
Now, to determine if the force is attractive or repulsive, we need to look at the signs of the charges. In this case, both charges are positive, so the force between them is repulsive.