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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 - CBSE (Science) Class 12 - Physics

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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

Capacitance between the parallel plates of the capacitor, C = 8 pF

Initially, distance between the parallel plates was and it was filled with air. Dielectric constant of air, k= 1

Capacitance, C, is given by the formula,

`C=(kin_0A)/d`

`=(in_0A)/d`  ...(1)

Where,

A = Area of each plate

`in_0`= Permittivity of free space

If distance between the plates is reduced to half, then new distance, d = `d/2`

Dielectric constant of the substance filled in between the plates, k' = 6

Hence, capacitance of the capacitor becomes

`C'=(k'in_0A)/d=(6in_0A)/(d/2)`    ...(2)

Taking ratios of equations (i) and (ii), we obtain

`C'=2xx6C`

=12C

=12 x 8=96 pF

Therefore, the capacitance between the plates is 96 pF.

  Is there an error in this question or solution?

APPEARS IN

 NCERT Solution for Physics Textbook for Class 12 (2018 to Current)
Chapter 2: Electrostatic Potential and Capacitance
Q: 5 | Page no. 87
Solution 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 Concept: The Parallel Plate Capacitor.
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