Question

$\huge\underbrace\bold\red{Question :-}$
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?

${\rm # Sanvi}$
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Answer 1

Answer:

If C be the air filled capacitance , dielectric filled capacitance becomes C

′

=KC where K is the dielectric constant.

Initial charge , q

i

=CV

i

=500C and final charge q

f

=C

′

V

f

=75KC

As the system is isolated so charge will be constant.

Thus, q

i

=q

f

⇒500C=75KC⇒K=

75

500

=20/3

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Answer 2

Answer:

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? Therefore, the capacitance between the plates is 96 pF.

Explanation: