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Energy Stored in a Capacitor A capacitor is basically a two terminal electrical component used for storage of energy electrostatically in an electric field. A large variety of capacitors are available but one thing that is common among all is that each capacitor contains at least two electrical conductors which are separated or partitioned by a dielectric insulator. One of the most commonly used capacitor consists of metal foils with a thin layer of insulating film in between. An electric field develops across the dielectric as a result of potential difference across the conductors. This causes both the charges to separate and the positive charge gets accumulated on one plate and the negative charge on the other. The electrostatic field stores the energy. The most appropriate capacitor is the one which has a single constant value i.e. the capacitance and it is defined as the quotient of electric charge on each conductor to the potential difference between them. The capacitance is at its peak when there is a thin separation between the big areas of conductor and it is due to this that the capacitor conductors are also called plates. Now we discuss some formulae and results based on the energy stored in capacitor: We consider the case of charging a parallel plate capacitor which is originally uncharged. We shift a charge Q from one plate to another thus resulting in the former plate having a charge -Q and the later with charge +Q. once some electric charge has been transferred it leads to the development of some electric field which prohibits the transfer of any further charge. Hence, if we wish to charge the capacitor completely we need to work against this field and this work performed will constitute the energy stored in the capacitor. Let us see how this energy is calculated: Consider the figure given above. We assume that the capacitor plates carry a charge q while the potential difference between the plates is V. hence, the work done in shifting the infinitesimal amount of charge dq from the negative plate to the positive plate is given by dW = V dq Now we wish to calculate the total work W(Q) performed in shifting of total charge Q from one plate to another. For finding it, we can divide the total charge into various small increments dq and then find the incremental work dW done in shifting of this incremental charge using the above listed formula. The only issue one might face in this is that the potential difference V between the plates being a function of total charge transmitted and so, V (q) = q/C dW = q dq/ C On integrating we obtain Now, again we know that the work done in charging the capacitor equals the energy stored in the capacitor. Also since we know that C = Q/V, so the stored energy can be represented in any of the following equivalent forms: U_{C} = Q^{2}/2C = ½ CV^{2} = ½ Q.V Energy supplied by Source U_{S} = Q.V Energy lost in form of heat = ½ Q.V Energy density of the electric field= Force on a Dielectric in a Capacitor View the following video for more on capacitors Example: A 4μf capacitor is charged to 150 V and another 6μf capacitor is charged to 200 V. Then they are connected across each other. Find the potential difference across them. Calculate the heat produced. Solution: 4μf charged to 150 V would have q_{1} = C_{1}V_{1} = 600μC 6μf charged to 200 V would have q_{2} = C_{2}V_{2} = 1200μC After connecting them across each other, they will have a common potential difference V. Charges will readjust as q_{1}’ and q_{2}’ Some of the key points on storage of energy in a capacitor:
A capacitor is basically a two terminal electrical component used for storage of energy electrostatically in an electric field. A large variety of capacitors are available but one thing that is common among all is that each capacitor contains at least two electrical conductors which are separated or partitioned by a dielectric insulator. One of the most commonly used capacitor consists of metal foils with a thin layer of insulating film in between.
An electric field develops across the dielectric as a result of potential difference across the conductors. This causes both the charges to separate and the positive charge gets accumulated on one plate and the negative charge on the other. The electrostatic field stores the energy. The most appropriate capacitor is the one which has a single constant value i.e. the capacitance and it is defined as the quotient of electric charge on each conductor to the potential difference between them.
The capacitance is at its peak when there is a thin separation between the big areas of conductor and it is due to this that the capacitor conductors are also called plates.
Now we discuss some formulae and results based on the energy stored in capacitor:
We consider the case of charging a parallel plate capacitor which is originally uncharged. We shift a charge Q from one plate to another thus resulting in the former plate having a charge -Q and the later with charge +Q. once some electric charge has been transferred it leads to the development of some electric field which prohibits the transfer of any further charge. Hence, if we wish to charge the capacitor completely we need to work against this field and this work performed will constitute the energy stored in the capacitor. Let us see how this energy is calculated:
Consider the figure given above. We assume that the capacitor plates carry a charge q while the potential difference between the plates is V. hence, the work done in shifting the infinitesimal amount of charge dq from the negative plate to the positive plate is given by
dW = V dq
Now we wish to calculate the total work W(Q) performed in shifting of total charge Q from one plate to another. For finding it, we can divide the total charge into various small increments dq and then find the incremental work dW done in shifting of this incremental charge using the above listed formula.
The only issue one might face in this is that the potential difference V between the plates being a function of total charge transmitted and so, V (q) = q/C
dW = q dq/ C
On integrating we obtain
Now, again we know that the work done in charging the capacitor equals the energy stored in the capacitor. Also since we know that C = Q/V, so the stored energy can be represented in any of the following equivalent forms:
U_{C} = Q^{2}/2C = ½ CV^{2} = ½ Q.V
Energy supplied by Source
U_{S} = Q.V Energy lost in form of heat = ½ Q.V
Energy density of the electric field= Force on a Dielectric in a Capacitor View the following video for more on capacitors
Example:
A 4μf capacitor is charged to 150 V and another 6μf capacitor is charged to 200 V. Then they are connected across each other. Find the potential difference across them. Calculate the heat produced.
Solution: 4μf charged to 150 V would have q_{1} = C_{1}V_{1} = 600μC 6μf charged to 200 V would have q_{2} = C_{2}V_{2} = 1200μC After connecting them across each other, they will have a common potential difference V.
Charges will readjust as q_{1}’ and q_{2}’
Some of the key points on storage of energy in a capacitor:
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