As shown in the figure, two parallel plate capacitors having equal plate area of 200 cm² are joined in such a way that a ≠ b. The equivalent capacitance of the combination is x ∈₀ F. The value of x is ____________.

Solution

Physics sir · 4 steps

The two middle plates are wired together, so the 1 mm gap between them does no work; only a + b counts.

1. Read the circuit. The second and third plates are joined by a wire, so they act as one conductor. That leaves two capacitors in series: one with gap a and one with gap b, both with area A.
2. Series adds gaps. For two series capacitors of equal area, 1/C = a/(ε₀ A) + b/(ε₀ A). So the combination behaves like one capacitor with gap a + b. The values of a and b separately do not matter.

   C = (ε₀ A)/(a+b) = (ε₀ A)/(d-c)

3. Put in numbers. a + b = 5 - 1 = 4 mm = 4 × 10⁻³ m and A = 200 cm² = 2 × 10⁻² m². So C = ε₀ × (2 × 10⁻²)/(4 × 10⁻³) = 5ε₀.
4. So. The equivalent capacitance is 5ε₀ F, so x = 5.

The common slip: 4. Using the full d = 5 mm as the gap gives (2 × 10⁻²)/(5 × 10⁻³) = 4. The 1 mm between the joined plates must be taken out first.

The idea to keep. Two plates joined by a wire are one conductor, and the gap between them carries no field. Equal-area capacitors in series act like a single capacitor whose gap is the sum of the air gaps.

Answer from the NTA final answer key. Solution by PrepWiser.