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A charge of 8.15 nC is located 1.69 m from a 4.26 nC point charge.

(a) Find the magnitude of the electrostatic force that one charge exerts on the other.
[tex] \text{N} [/tex]

(b) Is the force attractive or repulsive?
A. Attractive
B. Repulsive

Answer :

Final answer:

The force between the two charges would be approximately 1.15x10^-8 N according to Coulomb's Law. As both charges are positive, they would repel each other, so the force would be repulsive.

Explanation:

The force between two charged particles can be found using Coulomb's Law, which states that the force (F) between two charges is proportional to the product of the charges (q1 and q2), and inversely proportional to the square of the distance (r) between them. The equation for Coulomb's Law is F = k*q1*q2/r². Here, k is Coulomb's constant (approximately 8.99x10^9 Nm²/C² in a vacuum).

Applying this to your question, the formula will be F = (8.99x10^9 Nm²/C² * (8.15x10^-9 C) * (4.26x10^-9 C)) / (1.69 m)². After calculating, you should find the magnitude of the force to be approximately 1.15x10^-8 N.

As for whether the force is attractive or repulsive, remember that like charges repel and unlike charges attract. Since both charges are given as positive (there is no negative sign), these charges would repel each other, so the force is repulsive.

Learn more about Coulomb's Law here:

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