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Chapter 16: Problem 13

\(\mathrm{A}+2.0\) -nC point charge is \(3.0 \mathrm{cm}\) away from a $-3.0 \mathrm{-nC}$ point charge. (a) What are the magnitude and direction of the electric force acting on the +2.0 -nC charge? (b) What are the magnitude and direction of the electric force acting on the -3.0 -nC charge?

Short Answer

Expert verified
(b) What is the magnitude and direction of the electric force acting on the -3.0 nC charge? Answer: (a) For the +2.0 nC charge: Magnitude of the electric force is 6.0 x 10^-9 N and the direction is towards the -3.0 nC charge. (b) For the -3.0 nC charge: Magnitude of the electric force is 6.0 x 10^-9 N and the direction is towards the +2.0 nC charge.

Step by step solution

01

Coulomb's Law formula

Write down the formula for Coulomb's Law: \(F = k \frac{q_1 q_2}{r^2}\) where \(F\) is the force between the two charges, \(k\) is Coulomb's constant (\(8.99 \times 10^9 \mathrm{Nm^2/C^2}\)), \(q_1\) and \(q_2\) are the magnitudes of the two charges, and \(r\) is the distance between the charges.
02

Convert distances and charges to appropriate units

Convert the distances and charges to appropriate units (meters and coulombs). In our case, we have: \(q_1 = +2.0 \text{nC} = +2.0 \times 10^{-9} \mathrm{C}\) \(q_2 = -3.0 \text{nC} = -3.0 \times 10^{-9} \mathrm{C}\) \(r = 3.0 \mathrm{cm} = 0.03 \mathrm{m}\)
03

Calculate electric force acting on +2.0 nC charge

Plug the values from step 2 into the Coulomb's Law formula, to find the electric force acting on the +2.0 nC charge: \(F_{+2} = k \frac{(+2.0 \times 10^{-9} \mathrm{C})(-3.0 \times 10^{-9} \mathrm{C})}{(0.03 \mathrm{m})^2}\) \(F_{+2} = (8.99 \times 10^9 \mathrm{Nm^2/C^2}) \frac{(-6.0 \times 10^{-18} \mathrm{C^2})}{(0.0009 \mathrm{m^2})}\) \(F_{+2} = -6.0 \times 10^{-9} \mathrm{N}\) The force acting on the +2.0 nC charge is \(6.0 \times 10^{-9} \mathrm{N}\) towards the -3.0 nC charge (as the force is negative).
04

Calculate electric force acting on -3.0 nC charge

From Newton's Third Law, we know that the forces in opposite directions are equal in magnitudes, so: \(F_{-3} = -F_{+2}\) Therefore, the force acting on the -3.0 nC charge is also \(6.0 \times 10^{-9} \mathrm{N}\), but its direction is towards the +2.0 nC charge (as the force is positive).
05

Final Answers

(a) For the +2.0 nC charge: Magnitude of the electric force is \(6.0 \times 10^{-9} \mathrm{N}\) and the direction is towards the -3.0 nC charge. (b) For the -3.0 nC charge: Magnitude of the electric force is \(6.0 \times 10^{-9} \mathrm{N}\) and the direction is towards the +2.0 nC charge.

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