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Question: The force between 2 charges is 50N and they are separated by a distance of 0.04 meters. One charge is 0.02C, what is the other charge?
To find the other charge, we can use Coulomb's Law, which states that the force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \). Given: - \( F = 50 \, \text{N} \) - \( q_1 = 0.02 \, \text{C} \) - \( r = 0.04 \, \text{m} \) We need to find \( q_2 \). Rearranging Coulomb's Law to solve for \( q_2 \), we have: \[ q_2 = \frac{F \cdot r^2}{k \cdot q_1} \] Substituting the given values into the equation, we get: \[ q_2 = \frac{50 \, \text{N} \times (0.04 \, \text{m})^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \times 0.02 \, \text{C}} \] Calculating the values: \[ q_2 = \frac{50 \times 0.0016}{8.99 \times 10^9 \times 0.02} \] \[ q_2 = \frac{0.08}{1.798 \times 10^8} \] \[ q_2 \approx 4.45 \times 10^{-10} \, \text{C} \] So the other charge \( q_2 \) is approximately \( 4.45 \times 10^{-10} \, \text{C} \).
We can use Coulomb's Law to solve this problem. Coulomb's law is stated as F = k * q1 * q2 / r^2, where: - F is the force between the charges (in newtons) - k is Coulomb's constant, approximately 8.99 * 10^9 N m^2/C^2 - q1 and q2 are the magnitudes of the charges (in coulombs) - r is the distance between the charges (in meters) In this case, you want to find q2 (let's call this Q), the magnitude of the second charge. Rearranging the equation to solve for Q gives us Q = F * r^2 / (k * q1). Substituting the given values into the equation: Q = (50 N * (0.04 m)^2) / ((8.99 * 10^9 N m^2/C^2) * 0.02 C) Compute the values carefully and you will find the value of Q, the magnitude of the second charge.
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