Electrostatics – Net Electrostatic Force on Charge Q
Question:
Question 2:
Points A and B lie along the line joining two equal and opposite charges +q and −q. Each point is a distance b from one of the charges, and the total distance between the charges is d = 4b.
Suppose a third charge Q = 27q is placed at point B. What is the magnitude of the electrostatic force on charge Q in terms of kq² / b²?
Choose the correct option:
a) 20 b) 30 c) 40 d) 50 e) 60
Answer:
Step 1: Given Information
- Charge at A: +q
- Charge at some midpoint: −q
- Distance between +q and −q: d = 4b
- So, point B is located at b away from −q, and 3b from +q
- Test charge placed at point B: Q = 27q
Step 2: Apply Coulomb’s Law for Each Force
The electrostatic force between two charges is given by:
F = k × |q₁q₂| / r²
Force on Q due to +q (distance = 3b):
F₁ = k × (27q × q) / (3b)² = 27kq² / 9b² = 3kq² / b² → Towards right
Force on Q due to −q (distance = b):
F₂ = k × (27q × q) / b² = 27kq² / b² → Towards right
Step 3: Net Force on Charge Q
Both forces are towards the same direction (right), so they add up:
Fnet = F₁ + F₂ = 3kq² / b² + 27kq² / b² = 30kq² / b²
The magnitude of the electrostatic force on charge Q is:
30 × (kq² / b²)
Correct Option: (b) 30