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Chapter 21: Problem 35

Two identically charged particles separated by a distance of \(1.00 \mathrm{~m}\) repel each other with a force of \(1.00 \mathrm{~N}\). What is the magnitude of the charges?

Short Answer

Expert verified
Answer: The magnitude of each charge is approximately \(3.33 \times 10^{-6} \ \mathrm{C}\).

Step by step solution

01

Write down the given variables and the formula for Coulomb's Law.

We are given: - Force, F = 1.00 N - Distance between the charges, r = 1.00 m - The charges are identical, q1 = q2 = q Coulomb's Law formula is: \(F = k \frac{q_1 q_2}{r^2}\) where - F is the force between two charges - k is Coulomb's constant (\( k \approx 8.99 \times 10^9 \ \mathrm{N m^2C^{-2}}\)) - q1 and q2 are the magnitudes of the charges - r is the distance between the centers of the charges
02

Plug the given values into the formula and solve for q.

We want to find the magnitude of the charges q. Since both charges are identical, we can write the equation as follows: \(F = k \frac{q^2}{r^2}\) \(1.00 \ \mathrm{N} = (8.99 \times 10^9 \ \mathrm{N m^2 C^{-2}}) \frac{q^2}{(1.00 \ \mathrm{m})^2}\)
03

Solve for q.

Now, we'll solve for the magnitude of the charge q: \(q^2 = \frac{1.00 \ \mathrm{N} \cdot (1.00 \ \mathrm{m})^2}{8.99 \times 10^9 \ \mathrm{N m^2 C^{-2}}}\) \(q^2 \approx 1.11 \times 10^{-10} \ \mathrm{C^2}\) \(q \approx \sqrt{1.11 \times 10^{-10} \ \mathrm{C^2}}\)
04

Calculate the final result.

Finally, we take the square root of the found value to determine the magnitude of the charges: \(q \approx 3.33 \times 10^{-6} \ \mathrm{C}\) So, the magnitude of each charge is approximately \(3.33 \times 10^{-6} \ \mathrm{C}\).

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Most popular questions from this chapter

Performing an experiment similar to Millikan's oil drop experiment, a student measures these charge magnitudes: \(3.26 \cdot 10^{-19} \mathrm{C} \quad 5.09 \cdot 10^{-19} \mathrm{C} \quad 1.53 \cdot 10^{-19} \mathrm{C}\) \(6.39 \cdot 10^{-19} \mathrm{C} \quad 4.66 \cdot 10^{-19} \mathrm{C}\) Find the charge on the electron using these measurements.

Two cylindrical glass beads each of mass \(m=10.0 \mathrm{mg}\) are set on their flat ends on a horizontal insulating surface separated by a distance \(d=2.00 \mathrm{~cm} .\) The coefficient of static friction between the beads and the surface is \(\mu_{\mathrm{s}}=0.200 .\) The beads are then given identical charges (magnitude and sign). What is the minimum charge needed to start the beads moving?

Four point charges are placed at the following \(x y\) coordinates: \(Q_{1}=-1 \mathrm{mC},\) at \((-3 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{2}=-1 \mathrm{mC},\) at \((+3 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{3}=+1.024 \mathrm{mC},\) at \((0 \mathrm{~cm}, 0 \mathrm{~cm})\) \(Q_{4}=+2 \mathrm{mC},\) at \((0 \mathrm{~cm},-4 \mathrm{~cm})\) Calculate the net force on charge \(Q_{4}\) due to charges \(Q_{1}, Q_{2}\) and \(Q_{3}\).

In general, astronomical objects are not exactly electrically neutral. Suppose the Earth and the Moon each carry a charge of \(-1.00 \cdot 10^{6} \mathrm{C}\) (this is approximately correct; a more precise value is identified in Chapter 22 ). a) Compare the resulting electrostatic repulsion with the gravitational attraction between the Moon and the Earth. Look up any necessary data. b) What effects does this electrostatic force have on the size, shape, and stability of the Moon's orbit around the Earth?

A metal plate is connected by a conductor to a ground through a switch. The switch is initially closed. A charge \(+Q\) is brought close to the plate without touching it, and then the switch is opened. After the switch is opened, the charge \(+Q\) is removed. What is the charge on the plate then? a) The plate is uncharged. b) The plate is positively charged. c) The plate is negatively charged. d) The plate could be either positively or negatively charged, depending on the charge it had before \(+Q\) was brought near.
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