Question

When an air capacitor with a capacitance of 340 nF (1 nF = 10−9F) is connected to a power supply, the energy stored in the capacitor is 1.65×10⁻⁵ J . While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 2.30×10⁻⁵ J

What is the potential difference between the capacitor plates?

What is the dielectric constant of the slab?

Two capacitors are connected parallel to each other and connected to the battery with voltage 25 V . Let C1 = 9.9 μF ,C2 = 4.1 μF be their capacitances. Suppose the charged capacitors are disconnected from the source and from each other, and then reconnected to each other with plates of opposite sign together.

By how much does the energy of the system decrease?

Answer 1

Solution 1)

Air capacitor's capacitance = 340nF =340*10^-9 F

Energy stored in the capacitor = Ui =(1/2)CV^2

Ui= 1.65×10−5J .

(1/2)CV^2= Ui

So, V = sq rt (2Ui/C)

V = sq rt ( 3.3x10^-5/ 340x10^-9 )

V = 9.85V (Ans)

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Now, increased energy ,Uf =1.65×10−5J + 2.30×10−5 J.

Uf =3.95×10−5 J

We know,
Uf=(1/2)(dielectric constant)CV^2

Also,
dielectric constant = Uf / Ui

dielectric const =3.95×10−5 /1.65×10−5

dielectric constant =2.39 (Ans)

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Good luck!:)