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Post

Electric Force Between Two Charges in Vacuum

Created by @nathanedwards

at November 1st 2023, 8:08:27 am.

AP_Physics_1-Questions

#charges

#coulombs-law

#electric-force

Question:

Two point charges, Q₁ and Q₂, are placed 2 meters apart in a vacuum. The charge Q₁ has a magnitude of 4 microcoulombs and the charge Q₂ has a magnitude of 3 microcoulombs. Calculate the electric force between the two charges.

Answer:

The electric force between two charges can be calculated using Coulomb's Law:

F_e = \frac{{k \cdot |Q_1| \cdot |Q_2|}}{{r^2}}

Where: F_e is the electric force, k is the electrostatic constant (8.99 × 10^9 N m²/C²), |Q₁| and |Q₂| are the magnitude of the charges, and r is the distance between the charges.

Given: |Q₁| = 4 μC = 4 × 10^(-6) C |Q₂| = 3 μC = 3 × 10^(-6) C r = 2 m

Substituting the given values into Coulomb's Law:

F_e = \frac{{(8.99 \times 10^9) \cdot (4 \times 10^{-6}) \cdot (3 \times 10^{-6})}}{{2^2}}

Simplifying:

F_e = \frac{{8.99 \times 4 \times 3 \times 10^{-6} \times 10^{-6}}}{{4}}

F_e = (8.99 \times 4 \times 3) \times (10^{(-6+(-6))}) \times \frac{1}{4}

F_e = 107.88 \times 10^{-12} \times \frac{1}{4}

F_e = 26.97 \times 10^{-12} N

Therefore, the electric force between the two charges is approximately 26.97 piconewtons (pN).