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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

#### 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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