Question
3. Three point charges are placed at the following points on the x-axis: +2.0mu Calx=0 at x=40.0 cm, and -5.0mu C at x=120.0 cm. Determine the force on the -3.0mu C charge. Iiil [ans: 0.55 N toward the negative x-direction or 0.55 N [left]]
Solution
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(295 Votes)
Uriah
Master · Tutor for 5 years
Answer
Here's how to determine the force on the -3.0 μC charge:**1. Define Variables and Constants:*** q₁ = +2.0 μC = +2.0 x 10⁻⁶ C (charge at x = 0)* q₂ = -3.0 μC = -3.0 x 10⁻⁶ C (charge at x = 40.0 cm = 0.40 m)* q₃ = -5.0 μC = -5.0 x 10⁻⁶ C (charge at x = 120.0 cm = 1.20 m)* k = 8.99 x 10⁹ Nm²/C² (Coulomb's constant)* r₁₂ = 0.40 m (distance between q₁ and q₂)* r₂₃ = 1.20 m - 0.40 m = 0.80 m (distance between q₂ and q₃)**2. Calculate the Force between q₁ and q₂ (F₁₂):**The force F₁₂ is attractive since the charges have opposite signs. Its direction is along the positive x-axis (to the right).F₁₂ = (k * |q₁| * |q₂|) / r₁₂²F₁₂ = (8.99 x 10⁹ Nm²/C² * 2.0 x 10⁻⁶ C * 3.0 x 10⁻⁶ C) / (0.40 m)²F₁₂ = 0.337 N (to the right)**3. Calculate the Force between q₂ and q₃ (F₂₃):**The force F₂₃ is repulsive since the charges have the same signs. Its direction is along the negative x-axis (to the left).F₂₃ = (k * |q₂| * |q₃|) / r₂₃²F₂₃ = (8.99 x 10⁹ Nm²/C² * 3.0 x 10⁻⁶ C * 5.0 x 10⁻⁶ C) / (0.80 m)²F₂₃ = 0.212 N (to the left)**4. Calculate the Net Force on q₂:**The net force on q₂ is the vector sum of F₁₂ and F₂₃. Since they act along the x-axis, we can simply subtract their magnitudes, considering their directions:F_net = F₂₃ - F₁₂ (Since F₂₃ is to the left and F₁₂ is to the right)F_net = 0.212 N - 0.337 NF_net = -0.125 N**5. Interpret the Result:**The negative sign indicates that the net force on the -3.0 μC charge is directed towards the negative x-direction (to the left) with a magnitude of 0.125 N.**Important Note:** The provided answer (0.55 N) is incorrect based on the given charges and distances. The correct answer, following the principles of Coulomb's Law, is 0.125 N to the left. Double-check the problem statement to ensure all values are correctly transcribed.