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  • A capacitor of is charged by a battery of 12 V. The battery is disconnected and a dielectric slab of dielectric constant 8 is inserted in between the plates of the capa

A capacitor of 4 \mu F is charged by a battery of 12 V. The battery is disconnected and a dielectric slab of dielectric constant 8 is inserted in between the plates of the capacitor to fill the space completely. Find the change in the (a) charge stored in the capacitor, (b) potential difference between the plates of the capacitor, and (c) energy stored in the capacitor.

 

Answers (1)

a)

\\\text{initial charge}\ Q=CV=4\mu F\times12V=48\mu C

Final charge=initial charge, therefore the change=0

b) Potential become V/8=12/8=1.5 volt

Change in potential =12-1.5=10.5Volts

c)

Change in energy

=\frac{1}{2}\times KC(V/8)^2-\frac{1}{2}\times C(V)^2

\\=\frac{1}{2}\times C(V)^2-\frac{1}{2}\times 8\times C(V/8)^2\\\\=\frac{7}{16}CV^2=\frac{7}{16}\times4\times10^{-6}\times144=252\mu J

Posted by

Safeer PP

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