Physics, asked by dhanagopal1706, 3 months ago

The air capacitor is charged to 0/1 1200 V and then filled with dielectric of K = 3. The charge on the plates will be [ Given - A = 150 sq cm and d = 0.8 mm]​

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Answered by ratansen
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The area of each plate of para...

The area of each plate of parallel plate air capacitor is 150cm2. The distance between its plates is 0.8 mm. It is charged to a pot. Diff of 1200 volt. What will be its energy ? What wil be the energy when it is filled with a medium of K = 3 and then charged. If it is charged. If it is charged first as an air capacitor and then filled with this dielectric, what will happen to energy ?

Updated On: 27-06-2022

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Solution

Here, A=150cm2=150×10−4m2,

d=0.8mm=8×10−4m,V0=1200V,C0=?E0=?

C0=∈0Ad=8.85×10−12×150×10−48×10−4

=1.66×10−10F

U0=12C0V20×1.66×10−10×(1200)2

=1.2×10−4J

With dielectric medium, capacity becomes

C=KC0=3×1.66×10−10farad

When charged to same potential , V=1200V,

Energy, U=12CV2=12(KC0)V20=K(E0)

=3×1.2×10−4=3.6×10−4J

When capacitor charged first as air capacitor, then on filling with the dielectric, its potential

becomes V=V0K=12003=400 volt, because capacity becomes 3 times whereas charge remain the same.

∴ New energy of capacitor

=12CV2=12(KC0)(V0K)2

=12C0V20K=1.2×10−43=4×10−5 joule

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