Question

Answer the question which is in attachment.

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Answer 1

Answer:

Electrostatic energy stored is given by

E = $\frac{1}{2}C V^{2}$ = $\frac{1}{2} X 600 X 10^{-12} X 200 X 200$ =12 X $10^{-6}$ J

Now, Charge on the capacitor,

Q = C V = 600 X $10^{-12}$ X 200 = 12 X $10^{-8}$

After addition with the other uncharged capacitor in parallel, their equivalent capacitance will be

C = C₁ + C₂ = 1200pF = 12 X $10^{-10}$ F

Now, energy stored will be given by:

E = $\frac{Q^{2} }{2C}$    

(NOTE - Here, overall charge will remain constant, as charge can neither be created nor be destroyed. So, I have converted the equation in the form of charge.)

or, E = $\frac{12 X 10^{-8} X 12 X 10^{-8} }{2 X 12 X 10^{-10} }$ = 6 X $10^{-6}$J

Answer 2

Question :

A 600pF capacitor is charged by a 200V supply. Calculate the electrostatic energy stored in it. It is then disconnected from the supply and connected in parallel to another uncharged 600pF capacitor. What is the energy stored in the combination ?

Solution :

Okay so we need to calculate the electrostatic energy first , of a single capacitor .

Capacitance is given as 600pF ,  μ is 10^(-12) .

It is charged with a 200V supply( which is the potential difference here )

Electrostatic Energy: $1/2$ cv² = 1/2 x 600 x 10^(-12) x 200 x 200

= 12 x 10^6 x 10^(-12)

=12 x 10^(-6) Joules

Now the capacitor is disconnected and then connected in parallel to another uncharged 600pF capacitor.

Lets calculate the equivalent capacitance first

Its 1200pF ( When in parallel, capacitance's add up )

Electrostatic energy for this case:

> 1/2 x 1200 x 10^(-12) x 200 x 200

> 24 x 10^6 x 10^(-12)

> 24 x 10^(-6) Joules

Answer :

(i) Electrostatic energy of the initial combination : 12*10^(-6)Joules

(ii) Electrostatic energy of the second combination : 24*10^(-6) Joules

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