Capacitor: charge and energy calculations

[hibphys]

Homework Statement

A capacitor of capacitance 12*10^6 is fully charged from a 20V d.c. supply.
i Calculate the charge stored by the capacitor.
ii calculate the energy delivered by the dc supply
iii calculate the energy stored in the capacitor.
iv account for the difference between your answers for ii and iii.

Homework Equations

i charge is Q=CV
ii I think this is W=QV
iii I think this is W=1/2 CV^2[/B]

The Attempt at a Solution

iv I am not so sure it may be the fact that the dc current is equal to the voltage across the circuit or potential difference or something. I don't really get that.
i 2.4e-4
ii4.8e-3
iii2.4e-3

 

[jbriggs444]

Apparently this capacitor is actually $10^{-6}$ farads, not $10^6$ as written. With that understanding, your answers for i, ii and iii are correct.

For part iv, consider what is happening as the capacitor is halfway charged. The dc supply (ideally) is putting current out across a 20 V potential difference between its two leads. The capacitor (ideally) has a 10 V potential difference between its two leads. How can one account for the other 10 V of potential difference?

 

[hibphys]

LOST TO RESISTANCE ?

 

[jbriggs444]

Yes. You could elaborate on that answer by offering possibilities for where the resistance could exist. Thinking about conservation of energy, you could think about where the energy that is "lost to resistance" goes.

 

[HalfLostAndSearching]

I would like to clarify and provide a more thorough explanation of the concepts and calculations involved in this problem.

Firstly, a capacitor is an electronic component that is used to store electrical energy in the form of an electric charge. It is made up of two conductive plates separated by an insulating material, known as the dielectric. The capacitance of a capacitor is a measure of its ability to store charge and is represented by the symbol C. It is measured in units of Farads (F).

Now, let us look at the calculations involved in this problem.

i) The charge stored by the capacitor can be calculated using the formula Q=CV, where Q is the charge in coulombs, C is the capacitance in Farads, and V is the voltage in volts. In this case, the charge stored by the capacitor is 12*10^6 * 20 = 2.4 * 10^-4 coulombs.

ii) The energy delivered by the DC supply can be calculated using the formula W=QV, where W is the energy in joules, Q is the charge in coulombs, and V is the voltage in volts. In this case, the energy delivered by the DC supply is 2.4 * 10^-4 * 20 = 4.8 * 10^-3 joules.

iii) The energy stored in the capacitor can be calculated using the formula W=1/2 CV^2, where W is the energy in joules, C is the capacitance in Farads, and V is the voltage in volts. In this case, the energy stored in the capacitor is 1/2 * 12 * 10^6 * (20)^2 = 2.4 * 10^-3 joules.

iv) The difference between the answers for ii) and iii) is due to the fact that in the first calculation, we are considering the energy delivered by the DC supply to the capacitor, while in the second calculation, we are considering the energy stored in the capacitor itself. When a capacitor is charged, it stores energy in the form of an electric field between its plates. This energy is then released when the capacitor is discharged.

In conclusion, the difference between the two answers is due to the fact that the energy delivered by the DC supply is not fully stored in the capacitor. Some of it is lost in the form of heat or other