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QPages 198-203: Q68: HC Verma: Concepts of Physics - Volume 2 Electric Current in Conductors Pages 198-203: Q68 Solution: A parallel-plate capacitor has plate area , plate separation…

A parallel-plate capacitor has plate area 20 cm220 \text{ cm}^2, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ100 \text{ k}\Omega resistor. Find the energy of the capacitor 8.9 _s after the connections are made.

Step-by-Step Verified Solution

**Answer:** U=2.4times10−8U = 2.4 \\times 10^{-8} J. **Step-by-Step Explanation:** A=20A = 20 cm_, d=1.0d = 1.0 mm, K=5.0K = 5.0, varepsilon=6.0\\varepsilon = 6.0 V, R=100textkOmegaR = 100\\ \\text{k}\\Omega, t=8.9mutextst = 8.9\\ \\mu\\text{s}. C=Kepsilon0A/d=8.85times10−11C = K\\epsilon_0 A/d = 8.85 \\times 10^{-11} F. tau=RC=8.85times10−6\\tau = RC = 8.85 \\times 10^{-6} s. At t=8.9mutextst = 8.9\\ \\mu\\text{s}, VC=varepsilon(1−e−t/tau)V_C = \\varepsilon(1 - e^{-t/\\tau}). U=frac12CVC2U = \\frac{1}{2}CV_C^2 [citation:2].

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