Question 5

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?

Answer

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

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

Capacitance, C, is given by the formula,

Where, A = Area of each plate

= 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, = 6

Hence, capacitance of the capacitor becomes

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

Therefore, the capacitance between the plates is 96 pF.

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A point charge +10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)

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In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10

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Nikita Bhatia
2019-06-27 16:35:05

Thanks for this nswer

Swati
2019-05-10 22:20:27

Thanks a lotð

Swati
2019-05-10 22:20:02

Thanks a lotð

Pratiksha
2019-02-22 16:49:08

Thnk u very helpful n good answer

Thank you
2019-01-31 12:49:12

Thank you

Akansha
2019-01-10 12:12:49

Akansha

Amaira
2018-04-03 17:39:42

Very helpful answer thank u

prasad
2017-08-03 00:32:46

good answers

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