Question

Two charged particles are a distance of 1.72 m from each other. One of the particles has a charge of 7.03 nC, and the other has a charge of 4.02 nC.
(A) What is the magnitude (in N) of the electric force that one particle exerts on the other?
______N

(B) Is the force attractive or repulsive?

Answer 1

A. The magnitude (in N) of the electric force that one particle exerts on the other is 8.60×10⁻⁸ N

B. The force is repulsive

A. How to determine the magnitude of the electric force

From the question given above, the following data were obtained:

• Charge 1 (q₁) = 7.03 nC = 7.03×10¯⁹ C
• Charge 2 (q₂) = 4.02 nC = 4.02×10¯⁹ C
• Electric constant (K) = 9×10⁹ Nm²/C²
• Distance apart (r) = 1.72 m
• Force (F) =?

The magnitude of the electric force can be obtained by using the Coulomb's law equation as shown below:

F = Kq₁q₂ / r²

F = (9×10⁹ × 7.03×10¯⁹ × 4.02×10¯⁹) / (1.72)²

F = 8.60×10⁻⁸ N

B. How to determine whether the force is attractive or repulsive

From the question given, we were told that:

• Charge 1 (q₁) = 7.03 nC
• Charge 2 (q₂) = 4.02 nC

Since both charge are positive, then the force attraction between them is repulsive as like charges repels and unlike charges attracts

Learn more about Coulomb's law:

brainly.com/question/506926

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Answer 2

The magnitude of the electric force that one particle exerts on the other is 8.59 x $10^{-8}$ N and the force is attractive.

How is Coulomb's Law Stated ?

Coulomb law states that the attractive or repulsive force between two point charges is proportional to the product of the charges and inversely proportional to square of the distance between them. That is,

F = KQq / r²

Given that two charged particles are a distance of 1.72 m from each other. One of the particles has a charge of 7.03 nC, and the other has a charge of 4.02 nC.

Where

• K = constant of proportionality = 9 x $10^{9}$  Nm²/c²

• Q = 7.03 x $10^{-9}$ C

• q = 4.02 x $10^{-9}$ C

• r = 1.72 m

(A) The magnitude of the electric force that one particle exerts on the other will be

F = (9 x $10^{9}$ x 7.03 x $10^{-9}$ x 4.02 x $10^{-9}$) / 1.72²

F = 2.54 x $10^{-7}$ / 2.96

F = 8.59 x $10^{-8}$ N

(B) Since our answer is positive, the force is therefore attractive.

Therefore, the magnitude of the electric force that one particle exerts on the other is 8.59 x $10^{-8}$ N and the force is attractive.

Learn more about Electric Force here: brainly.com/question/25923373

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