Answer :

Given:

C_{1}=5 μF

V_{1}=24 V

To calculate the charge present on the capacitor, we use the formula

where,

c = capacitance of the capacitor and

v = voltage across the capacitor

For first capacitor, the stored charge q_{1} is given by

Similarly for second capacitor, the stored charge q_{2} is given by-

Given, C_{2}=6 μF and V_{2}=12

a) Energy stored in each capacitor-

Energy stored in a capacitor is given by

Where, v = applied voltage

C =capacitance

For capacitor C_{1,} energy stored is given by

=

Similarly, for capacitor C_{2,} energy stored is given by

b) New charges on the capacitors when the positive plate of the first capacitor is now connected to the negative plate of the second nd vice versa

The capacitors are connected as shown on the right hand side.

The positive of first capacitor is connected to the negative of the second capacitor.

So charge flows from positive of first capacitor to the negative of the second capacitor.

Then, the net charge for connected capacitors becomes

Now, let *V* be the common potential of the two capacitors

Since, charge is conserved, we know that electric charge can neither be created nor be destroyed, hence net charge is always conserved.

From the conservation of charge before and after connecting, we get, common voltage V

We know,

where v = applied voltage and C is the capacitance

Using above relation, the new charges becomes-

c) Loss of electrostatic energy during the process

Energy stored in a capacitor is given by

1)

Where, v = applied voltage

C =capacitance

For capacitor C_{1,} energy stored is given by

=

Similarly, for capacitor C_{2 ,} energy stored is given by

=

2)

But given

for c_{1}, actual V_{1} = 24V

and c_{2,} actualV_{2} = 12V

Using 1)

And

Now, change in energy,

3)

From 2) and3)

Loss of electrostatic energy =

d) This energy, which is lost as electrostatic energy gets converted and dissipated from the capacitor in the from of heat energy.

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