Question

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_____.

Answer 1

Correct Answer: 5

As per the arrangement given, the distance between the capacitor plates are \(a\) and \(b\) and \(a \neq b\). Using the diagram we can write \[ b = 5 - a - 1 = (4 - a) \text{ in mm} \] As we know the capacitance of a capacitor \(C = \dfrac{\varepsilon_0 A}{d}\) and in series arrangement \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \] \[ \frac{1}{C_{eq}} = \frac{a A}{\varepsilon_0 A} + \frac{4 - a}{\varepsilon_0 A} = \frac{4 \text{ (in mm)}}{\varepsilon_0 A} \] or \[ C_{eq} = \frac{\varepsilon_0 A}{4(\text{mm})} \] Given \(A = 200 \text{ cm}^2\), \[ C_{eq} = \varepsilon_0 \times 200 \times 10^{-4} = 4 \times 10^{-3} = 50 \times 10^{-1} \] or \[ C_{eq} = 5 \varepsilon_0 \text{ farad} \] Therefore, \(C_{eq} = 5 \varepsilon_0\), so \(x = 5\).