Question

# A parallel-plate capacitor of capacitance $$5 \mu F$$ is connected to a battery of emf $$6 V.$$ The separation between the plates is $$2 mm$$. A dielectric slab of thickness $$1 mm$$ and dielectric constant $$5$$ is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination.

Solution

## $$C = 5 \mu f \,\, \\V = 6 V \,\, \\d = 2mm = 2 \times 10^{-3} m$$$$d = 2 \times 10^{-3} m$$$$t = 1 \times 10^{-3} m$$$$k = 5$$ or                    $$C = \dfrac{\varepsilon_0 A}{d}$$$$\Rightarrow 5 \times 10^{-6} = \dfrac{8.85 \times A \times 10^{-12}}{2 \times 10^{-3}} \times 10^{-9}\\ \ \ \ \ \ \ \ \ \ \ \Rightarrow A = \dfrac{10^4}{8.85}$$When the dielectric placed on it$$C_1 = \dfrac{\varepsilon_0 A}{d - t + \dfrac{t}{k}} = \dfrac{8.85 \times 10^{-12} \times \dfrac{10^4}{8.85}}{2\times 10^{-3}-10^{-3} + \dfrac{10^{-3}}{5}} = \dfrac{10^{-12} \times 10^4 \times 5}{6 \times 10^{-3}} = \dfrac{5}{6} \times 10^{-5} = 0.00000833 = 8.33 \mu F$$Physics

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