Capacitor With Dielectrics: Practice Problem & Solution
The capacity of a parallel plate condenser is \(C_0\). If a dielectric of relative permittivity \(\varepsilon_r\) and thickness equal to one fourth the plate separation is placed between the plates, then its capacity becomes \(C\). The value of \(\frac{C}{C_0}\) will be :
Solution Explained:
To solve this problem, we apply the core principles of Capacitor With Dielectrics. Understanding the underlying formula is key to arriving at the correct answer below:
The initial capacitance is \(C_0 = \frac{\varepsilon_0 A}{d}\). With a dielectric slab of thickness \(t = d/4\), the new capacitance is \(C = frac{\varepsilon_0 A}{d - t + \frac{t}{\varepsilon_r}} = \frac{\varepsilon_0 A}{\frac{3d}{4} + \frac{d}{4\varepsilon_r}}\). Simplifying this gives \(C = C_0 \frac{4\varepsilon_r}{3\varepsilon_r + 1}\).
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