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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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