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Mademuasel [1]
2 years ago
12

Two positive point charges are 4.9cm apart. If the electric potential energy is 70.0 μJ, what is the magnitude of the force betw

een the two charges?
Physics
1 answer:
ExtremeBDS [4]2 years ago
6 0

Hi there!

Recall the following:

V \text{ (Electric Potential Energy) } = \frac{kq_1q_2}{r}\\\\F_E = \frac{kq_1q_2}{r^2}

k = Coulomb's Constant (Jm/C²)

q = Charge (C)
r = distance between charges (m)

To calculate the electric force between the two charges, we can simply divide by another 'r' (distance):

F_E = \frac{70}{0.049} = \boxed{1428.57 \mu J}

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