Charge A= 5uC and Charge B= 6uC were separated at a certain distance. How far they were from each other given that their electric potential energy is at 2 Joules.Group of answer choices2x10^11 m0.1 cm0.07 m0.1 m
Question
Charge A= 5uC and Charge B= 6uC were separated at a certain distance. How far they were from each other given that their electric potential energy is at 2 Joules.Group of answer choices2x10^11 m0.1 cm0.07 m0.1 m
Solution
The formula for electric potential energy (U) between two charges is given by:
U = k * q1 * q2 / r
where:
- U is the electric potential energy
- k is Coulomb's constant (8.99 x 10^9 N m^2/C^2)
- q1 and q2 are the charges
- r is the distance between the charges
Given in the problem:
- U = 2 J
- q1 = 5 uC = 5 x 10^-6 C (since 1 uC = 10^-6 C)
- q2 = 6 uC = 6 x 10^-6 C
- k = 8.99 x 10^9 N m^2/C^2
We can substitute these values into the formula and solve for r:
2 = (8.99 x 10^9) * (5 x 10^-6) * (6 x 10^-6) / r
Rearranging the equation to solve for r gives:
r = (8.99 x 10^9) * (5 x 10^-6) * (6 x 10^-6) / 2
Solving this equation will give the distance between the charges.
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