13. A 3.0 uF air capacitor is charged to a potential 12.0 V. A slab of dielectric
constant K = 7 is made to fill the space. Calculate the ratio of the energies
stored in the two systems.
Answers
Answer:
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Answer:
Situation 1- Energy of the capacitor of capacitance C=3μF, which is filled with the air and is charged to a potential 12V, is=E1= (1/2) Q^2/C,
Where Q is the charge on the plates of the capacitor
Situation 2- A slab of dielectric constant of 7 is inserted between the plates of the above capacitor and it replaces the air. In the process, the charge Q on the plates remains undisturbed . However, the electric field as well as the voltage difference decrease by a factor of 7. In other words, the capacitance increases to 7C or 21μF. The Energy of the capacitor of 21μF, which is filled with the slab of dielectric constant 7 and has a charge Q =
E2= (1/2) Q^2/7C
E1/E2 = 7
Explanation:
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