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Electrostatic Potential And Capacitance

Question

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10^–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Solution

Given, capacitance, C₁ = 8 pF

$C_{1} = \frac{ε_{o} A}{d} = 8 pF C_{2} = K \frac{ε_{o} A}{d / 2} = \frac{6 \times 2 ε_{o} A}{d} = 12 \times 8 = 96 pF$

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Electrostatic Potential And Capacitance

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10^–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

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Electrostatic Potential And Capacitance

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?

Some More Questions From Electrostatic Potential and Capacitance Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

A parallel plate capacitor with air between the plates has a capacitance of 8 pF (1pF = 10^–12 F). What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?