Parallel Plate Capacitor: Practice Problem & Solution
Assertion (A): If capacitor is filled with, same thickness \(t < d\) of dielectric and conducting sheet one after another, then capacitance are \(C_1\) and \(C_2\) respectively then \(C_1 < C_2\). Reason (R): Capacitance is more in presence of metal sheet in compare to dielectric sheet as
Short Solution (Best for NEET):
To solve this problem, we apply the core principles of Parallel Plate Capacitor. Understanding the underlying formula is key to arriving at the correct answer below:
Assertion (A): For a dielectric slab of thickness \(t\) and dielectric constant \(K\), \(C_1 = \frac{\epsilon_0 A}{d-t+t/K}\). For a conducting slab of thickness \(t\), \(C_2 = \frac{\epsilon_0 A}{d-t}\). Since \(K>1\), \(d-t+t/K > d-t\), implying \(C_1 < C_2\). So (A) is true.
Reason (R): A metal (conductor) effectively acts as a dielectric with \(K = \infty\), which makes its capacitance higher than a dielectric with a finite \(K\). So (R) is true and correctly explains (A).
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