Two charges, q1=-11.5nC and q2=28.5nC, are separated by a distance d=3.00cm as shown in the figure. Determine the following. (a) the electric potential (in kV ) at point A kV (b) the electric potential (in kV ) at point B, which is halfway between the charges kV

Answer

🔷 Electric Potential Due to Two Point Charges

🧪 Concept Overview

The electric potential at a point due to a point charge is a scalar quantity. It represents the electric potential energy per unit charge at that point.

• V = Electric potential in volts (V)
• k = Coulomb’s constant = 8.99 × 10⁹ Nm²/C²
• q = Charge in coulombs (C)
• r = Distance from the charge to the point (in meters)

📌 Given:

• q₁ = –11.5 nC = –11.5 × 10⁻⁹ C
• q₂ = 28.5 nC = 28.5 × 10⁻⁹ C
• Distance from each charge to Point A: d = 3.00 cm = 0.0300 m

🔹 Electric Potential at Point A

Using the principle of superposition:

Substituting values:

• V_A1 = (8.99 × 10⁹ × –11.5 × 10⁻⁹) / 0.0300 = –3446.2 V
• V_A2 = (8.99 × 10⁹ × 28.5 × 10⁻⁹) / 0.0300 = 8540.5 V

Total potential at Point A:

🔹 Electric Potential at Point B (Midpoint)

Point B is midway between the charges, so distance from each is 0.015 m.

Substituting values:

• V_B1 = (8.99 × 10⁹ × –11.5 × 10⁻⁹) / 0.0150 = –6892.3 V
• V_B2 = (8.99 × 10⁹ × 28.5 × 10⁻⁹) / 0.0150 = 17081.0 V

Total potential at Point B:

📊 Final Results

Point	Electric Potential (kV)
A	5.09
B	10.19

🧠 Interpretation

• Point A: Equal distance from both charges, but since one is negative and one is positive, their effects partially cancel. The result is a moderate net potential.
• Point B: Closer proximity to both charges increases the effect. Dominance of the larger positive charge results in a much higher potential.

📚 Concept Expansion

• Electric potential is a scalar quantity, so it’s added algebraically.
• A positive charge contributes a positive potential; a negative charge contributes a negative potential.
• The closer the point to a charge, the higher the magnitude of its contribution (due to the 1/r relationship).
• The superposition principle is a fundamental concept when multiple charges are involved.

🧪 Applications

• Electronics: Understanding voltage and capacitor behavior.
• Physics Labs: Mapping electric potentials in space.
• Biophysics: Analyzing neuron membrane potentials.
• Simulation Software: COMSOL, MATLAB, ANSYS for electrostatic field visualization.

✅ Conclusion

This problem clearly demonstrates the use of Coulomb’s law to determine electric potential in a two-charge system. The scalar nature of potential and its dependence on distance and charge sign are crucial to arriving at the correct values.

Final Answers:
V_A = 5.09 kV
V_B = 10.19 kV