Question

select the correct answer.
the electric force between a charged foam cup and a positively charged metal sphere is 2.12 newtons. if the charge on the cup is 2.0 × 10⁻⁶ coulombs and the charge on the metal sphere is 2.5 × 10⁻⁶ coulombs, what is the distance between the two? (k = 9.0 × 10⁹ newton·meter²/coulombs²)
a. 9.0 × 10⁻¹ meters
b. 6.4 × 10⁻¹ meters
c. 3.8 × 10⁻¹ meters
d. 2.8 × 10⁻¹ meters
e. 1.5 × 10⁻¹ meters

Explanation:

Step1: Recall Coulomb's Law

Coulomb's Law is given by \( F = k\frac{q_1q_2}{r^2} \), where \( F \) is the electric force, \( k \) is the electrostatic constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges. We need to solve for \( r \). Rearranging the formula for \( r \), we get \( r = \sqrt{k\frac{q_1q_2}{F}} \).

Step2: Substitute the given values

We are given \( F = 2.12 \, \text{N} \), \( q_1 = 2.0 \times 10^{-6} \, \text{C} \), \( q_2 = 2.5 \times 10^{-6} \, \text{C} \), and \( k = 9.0 \times 10^9 \, \text{N·m}^2/\text{C}^2 \). Substitute these values into the formula for \( r \):

First, calculate the product \( q_1q_2 \):
\( q_1q_2 = (2.0 \times 10^{-6} \, \text{C})(2.5 \times 10^{-6} \, \text{C}) = 5.0 \times 10^{-12} \, \text{C}^2 \)

Then, calculate \( k\frac{q_1q_2}{F} \):
\( k\frac{q_1q_2}{F} = (9.0 \times 10^9 \, \text{N·m}^2/\text{C}^2) \times \frac{5.0 \times 10^{-12} \, \text{C}^2}{2.12 \, \text{N}} \)
\( = \frac{9.0 \times 5.0 \times 10^{9 - 12}}{2.12} \, \text{m}^2 \)
\( = \frac{45 \times 10^{-3}}{2.12} \, \text{m}^2 \)
\( \approx \frac{0.045}{2.12} \, \text{m}^2 \)
\( \approx 0.021226 \, \text{m}^2 \)

Step3: Take the square root

Now, take the square root of the result to find \( r \):
\( r = \sqrt{0.021226 \, \text{m}^2} \approx 0.1457 \, \text{m} \)? Wait, that can't be right. Wait, let's recalculate the step \( k\frac{q_1q_2}{F} \) again.

Wait, \( 9.0 \times 10^9 \times 5.0 \times 10^{-12} = 9.0 \times 5.0 \times 10^{9 - 12} = 45 \times 10^{-3} = 0.045 \). Then divide by \( F = 2.12 \): \( 0.045 / 2.12 \approx 0.021226 \). Wait, but the square root of 0.021226 is approximately 0.1457, which is \( 1.457 \times 10^{-1} \, \text{m} \), close to option E. Wait, maybe I made a mistake in calculation. Wait, let's check again.

Wait, \( q_1 = 2.0 \times 10^{-6} \), \( q_2 = 2.5 \times 10^{-6} \), so \( q_1q_2 = 5.0 \times 10^{-12} \). Then \( kq_1q_2 = 9.0 \times 10^9 \times 5.0 \times 10^{-12} = 45 \times 10^{-3} = 0.045 \). Then \( 0.045 / 2.12 \approx 0.021226 \). Square root of 0.021226 is approximately 0.1457 meters, which is \( 1.457 \times 10^{-1} \, \text{m} \), which is approximately \( 1.5 \times 10^{-1} \, \text{m} \), which is option E.

Wait, maybe my initial calculation was correct. Let's verify:

\( r = \sqrt{\frac{kq_1q_2}{F}} = \sqrt{\frac{9.0 \times 10^9 \times 2.0 \times 10^{-6} \times 2.5 \times 10^{-6}}{2.12}} \)

Calculate numerator: \( 9.0 \times 2.0 \times 2.5 \times 10^{9 - 6 - 6} = 45 \times 10^{-3} = 0.045 \)

Then \( 0.045 / 2.12 \approx 0.021226 \)

Square root: \( \sqrt{0.021226} \approx 0.1457 \, \text{m} \approx 1.5 \times 10^{-1} \, \text{m} \)

Answer:

E. \( 1.5 \times 10^{-1} \) meters