At each of the four corners of a square of side a, a
charge +q is placed freely. What charge should be
placed at the centre of the square so that the whole
system be in equilibrium?
Answers
Answer:
- (2√2 + 1)q/4
Explanation:
Using, force(electrostatic) = kq₁q₂/r²
Here, q₁ = q₂ = q ; r = a and, r = a√2 for A-C(using Pythagoras theorem)
Thus,
Force by A = k qq/(a√2)² = k q²/2a²
Force by B = k qq/a² = k q²/a²
Force by D = k qq/a² = k q²/a²
Using vector addition, resultant
= √F(B)² + F(D)² + F(B).F(D)cos90
= √(kq²/a²)² + (kq²/a²)² + 0
= √2 (kq²/a²)
We add the resultant force and force by A, since both are in same line(direction). ∴ Opposing Force on C is
= √2 (kq²/a²) + (kq²/2a²)
= (2√2 + 1)kq²/2a²
Whereas, the charge at 0 must be provide same force to oppose the opposing force on C. Let the charge at O be Q.
distance between O and C is half of diagonal = a√2/a = a/√2
To be in equilibrium, force by O = - existing opposite force on C
∴ kqQ/(a/√2)² = - (2√2 + 1)kq²/2a²
⇒ Q/(a²/2) = - (2√2 + 1)q/2a²
⇒ 2Q = - (2√2 + 1)q/2
⇒ Q = - 1/2(2√2 + 1)q/2 = - (2√2 + 1)q/4
