Two balls have 3 nC of excess charge and are 10cm apart. What is the magnitude of the repulsive force on each?
Electric field = 27,000
Charge = 3e-8 Coulombs
F = EQ = 8.1e-4
...which is wrong.
use the electrostatic force equation
also, nano is E-9
I still get 7.1e-4 with this equation
f = k Q1 Q2 /r^2 = 9E9 * (3E-9)^2 / .1^2
F = 8.1E-6 N
QE = 9*10^9 *3*10^-9 *3*10^-9/.1^2
= 81*10^-9/10^-2
= 81*10^-7
= 8.1*10^-6 N
To calculate the magnitude of the repulsive force between two charged objects, you can use Coulomb's Law. Coulomb's Law states that the force between two charged objects is proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula is given as:
F = k * (q1 * q2) / r^2
Where:
F is the magnitude of the force
k is the electrostatic constant (approximately 9 x 10^9 Nm^2/C^2)
q1 and q2 are the charges of the objects
r is the distance between the objects
Given that you have two balls with 3 nC (3 x 10^-9 C) of excess charge and they are 10 cm (0.10 m) apart, you can calculate the force using Coulomb's Law.
First, convert the charge of each ball to Coulombs:
q = 3 x 10^-9 C
Next, substitute the values into the formula:
F = (9 x 10^9 Nm^2/C^2) * [(3 x 10^-9 C) * (3 x 10^-9 C)] / (0.10 m)^2
Simplifying the equation, you get:
F = 9 x 10^9 Nm^2/C^2 * 9 x 10^-18 C^2 / 0.01 m^2
F = (9 x 9) x (10^9 x 10^-18) / (0.01)^2
Calculating further:
F = 81 x (10^-9 x 10^-18) / 0.0001
F = 81 x 10^-27 / 0.0001
F = 81 x 10^-23 / 10^-4
F = 81 x 10^-19 N
Therefore, the magnitude of the repulsive force on each ball is approximately 8.1 x 10^-18 N.