Hii..
Can someone solve this please?
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Instead of a formula based approach, let me try to explain this conceptually.
1) A capacitor has two plates with opposite potential.
2) Opposite charges of equal magnitude, +Q and -Q accumulate on each plate.
3) Because of this, there is an electric field between the plates.
A capacitor stores its potential energy U, in this electric field. The more energy is stored, the denser this electric field gets.
What causes the field? The charges present on both plates. This field opposes the addition of more charges into the plates.
The electric field tries to push the plates back to their original state. It opposes the action of creating a charge imbalance.
The only thing preventing this is the voltage from the battery on both ends. Equilibrium.
If we do some work against this field, that work will be stored as potential energy inside the capacitor. So, to double the charges on both plates, how much charge do I have now?
Previously, we had +Q and -Q.
Now, we have +2Q and -2Q. Ignoring polarity, that's a total magnitude of 4Q.
So, the battery needs to do 4 times the amount of work to move those charges there. Thus, 4 times the amount of work gets stored there as potential energy in the electric field. The field is 4 times denser.
HOPE IT HELPS.
If you have any questions you can ask
1) A capacitor has two plates with opposite potential.
2) Opposite charges of equal magnitude, +Q and -Q accumulate on each plate.
3) Because of this, there is an electric field between the plates.
A capacitor stores its potential energy U, in this electric field. The more energy is stored, the denser this electric field gets.
What causes the field? The charges present on both plates. This field opposes the addition of more charges into the plates.
The electric field tries to push the plates back to their original state. It opposes the action of creating a charge imbalance.
The only thing preventing this is the voltage from the battery on both ends. Equilibrium.
If we do some work against this field, that work will be stored as potential energy inside the capacitor. So, to double the charges on both plates, how much charge do I have now?
Previously, we had +Q and -Q.
Now, we have +2Q and -2Q. Ignoring polarity, that's a total magnitude of 4Q.
So, the battery needs to do 4 times the amount of work to move those charges there. Thus, 4 times the amount of work gets stored there as potential energy in the electric field. The field is 4 times denser.
HOPE IT HELPS.
If you have any questions you can ask
Anonymous:
how did u conclude that work done wud be 4 times
as you mnow that W = (Q^2)/2C. So W∝(Q^2) [as the capacitance is same] So if the charge is doubled W∝ (2Q)^2 i.e. W ∝ 4Q^2 Hence the work done is 4 times
okay thanks
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