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Let us consider an ideal parallel plate capacitor in a circuit. The capacitor is connected to a battery which generates a potential difference of V across the terminals.

QUESTIONS:

1) The work required to charge the capacitor is calculated by assuming that a small charge dq is taken from the lower potential plate to the higher potential plate. Let us say that the potential difference between the plates at this point is q/C (C is the capacitance). When the charge dq goes through the battery it acquires a potential of V (assuming that the negative terminal is at 0 potential). The work done to take this charge from the high potential to low potential plate is q*dq/C. Where did the rest of the energy go ?

2) In order to do deposit the charge on the high potential plate some work is required to move the charge against the electric field of the capacitor. This work is said to be done by the battery. My question is : If the battery gives some electric potential energy to the charge then moving it to the high potential plate of the capacitor should increase the potential energy of the charge . Why is it then said that there is a potential drop across a capacitor ?

(EDIT : NEW QUESTIONS)

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3) While charging the capacitor in the above figure a charge dq is extracted from the negative plate. The charge gains a potential V after going through the battery. while approaching the high potential plate will the charge loose the Electric potential energy ?

4)The potential drop across a capacitor is Q/C. Let us assume that the emf of the battery is 5V. a charge goes through the battery and acquires some potential energy . Let us say that the potential drop across the capacitor is 2V. The potential of the charge after crossing the capacitor (displacing another charge on the low potential plate) will be 3V. What confuses me is the assumption that the high plate will be at 5V as well since only then the potential of the low potential plate can be said to be 3V. Where have I gone wrong?

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    $\begingroup$ Are you familiar with the two capacitor paradox? This sounds similar to it $\endgroup$ – Bob D Nov 10 at 8:14
  • $\begingroup$ @Bob D No , But I don't think it's relevant. My question concerns the potential drop across the capacitor. $\endgroup$ – Aditya Ahuja Nov 20 at 18:43
  • $\begingroup$ But you asked where the energy went. That's a separate part of the question, right?> $\endgroup$ – Bob D Nov 20 at 20:06
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Where did the rest of the energy go ?

It goes as [Ohmic/Joule heating](https://en.wikipedia.org/wiki/Joule_heating in any resistance which is in the circuit you question being similar to questions relating to the potential energy stored in a spring.

The electrical circuit is bound to have inductance and so you are dealing with an LCR circuit whose behaviour depends very much of the relative sizes of the values of the components which make up the circuit.
Usually when the charging of the capacitor is being considered the LCR system is considered to be over damped and the final steady state is reaches with no oscillation of charge in the system, ie a steady growth of the charge stored on the capacitor towards a final constant value when the voltage across the capacitor is equal to the emf of the battery.
If the resistance of the circuit is low enough then the system might be under damped and the final steady state, voltage across the capacitor equal to the emf of the battery is reached with the current in the circuit (and hence the charge on the capacitor) oscillating at the natural frequency of the LC(R) system.
The energy is still being lost due to Ohmic/Joule heating and eventually the system the steady state with half the energy delivered by the battery being stored in the capacitor and half dissipated as heat.

Energy can be lost from the system due to the emission of electromagnetic waves from the system although this effect usually contributes to insignificant amount of energy loss.
Whenever unbound charges (free electrons in this case) accelerate they emit em radiation. So if the current in the circuit is changing the circuit will emit em radiation however this effect is usually so small as to neglected. Very related to this is the capacitor paradox referred in a comment by @BobD.

Why is it then said that there is a potential drop across a capacitor ?

In this context potential drop means potential difference across the plates of the capacitor. So as the capacitor is charged the potential difference across the plates of the capacitor increases meaning that there is a greater drop in potential as one moves from the positive plate of the capacitor to the negative plate.

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  • $\begingroup$ But when we talk about potential drop we usually take in terms of charges right ? For example, A charge with some potential goes across a resistor and loses some potential energy. Is that not the case here? $\endgroup$ – Aditya Ahuja Nov 11 at 13:13
  • $\begingroup$ @AdityaAhuja whether you call it a potential drop or potential rise depends on which way you go around the circuit when applying KVL. In your circuit, if go around clockwise it is a potential drop across the capacitor. If you go counter-clockwise it's a potential rise. $\endgroup$ – Bob D Nov 20 at 19:14
  • $\begingroup$ @Bob D I'll rephrase.If the charge is not actually going across the capacitor and only displacing another charge why do we say that the potential difference is q/C. Isn't q/C the potential difference between the PLATES of the capacitor. Let me give an example of what I mean. Let us say a charge goes through a battery of 5 V (anode is the reference) and arrives at a capacitor plate. This charge displaces another charge . If the potential drop across the capacitor is 3 V at this time. Is it right to say that the displaced charge has a potential of 2 V? $\endgroup$ – Aditya Ahuja Nov 21 at 16:43
  • $\begingroup$ @AdityaAhuja First of all, I assume you understand that no charge "goes across" the capacitor if you mean it goes through the dielectric to get from one plate to the other. I also assume you also understand that when a battery moves charge (electrons) from one plate to the other those electrons arriving at the plate don't "displace" charges on that plate. The result is a net negative charge on the plate they are deposited on leaving an equal net positive charge, or a deficit of electrons, on the plate they were taken from. Is that how you understand it? $\endgroup$ – Bob D Nov 21 at 17:52
  • $\begingroup$ @Bob D The charge displaced is on the other plate. Also I know that the charge doesn't go across the dielectric. $\endgroup$ – Aditya Ahuja Nov 22 at 7:10
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1).....The work done to take this charge from the high potential to low potential plate is q*dq/C. Where did the rest of the energy go ??

The potential difference, or voltage, between two points is the work required per unit charge to move the charge between the points. So the work, or energy E, required by the battery to move charge Q between two points where the potential difference is V is

$$E_{battery}=QV=CV^2$$

Half of that energy delivered to the circuit by the battery is stored as electrical potential energy in the electric field of the capacitor or

$$E_{capacitor}=\frac{CV^2}{2}$$

The other half of the energy is dissipated as heat in the resistance in the circuit since work is required to move the charge through the resistance while the battery is charging (current > 0).

$$E_{resistor}=\frac{CV^2}{2}$$

2) In order to do deposit the charge on the high potential plate some work is required to move the charge against the electric field of the capacitor. This work is said to be done by the battery. My question is : If the battery gives some electric potential energy to the charge then moving it to the high potential plate of the capacitor should increase the potential energy of the charge . Why is it then said that there is a potential drop across a capacitor ?

While the battery is moving charge from one plate to the other it is increasing the potential difference, or voltage, between the plates until charging is complete and that voltage equals the battery terminal voltage. Whether you voltage across the capacitor a "potential drop" or "potential rise" depends on which way you go around the circuit when applying KVL. In your circuit, if go around clockwise it is a potential drop across the capacitor. If you go counter-clockwise it's a potential rise.

3) While charging the capacitor in the above figure a charge dq is extracted from the negative plate. The charge gains a potential V after going through the battery. while approaching the high potential plate will the charge loose the Electric potential energy ?

The charge loses potential energy flowing through the resistor and that loss of potential energy equals the heat dissipated in the resistor. Meanwhile the voltage across the capacitor is increasing meaning potential energy is being stored in the capacitor. In effect, part of the potential energy the battery gives the charge is lost in the resistor and the remaining potential energy of the charge is transferred to the capacitor.

4)The potential drop across a capacitor is Q/C. Let us assume that the emf of the battery is 5V. a charge goes through the battery and acquires some potential energy . Let us say that the potential drop across the capacitor is 2V. The potential of the charge after crossing the capacitor (displacing another charge on the low potential plate) will be 3V. What confuses me is the assumption that the high plate will be at 5V as well since only then the potential of the low potential plate can be said to be 3V. Where have I gone wrong?

I am having a hard time following this, but it seems you are ignoring the voltage drop across the resistor. If at some time the voltage across the capacitor is 2V then obviously at that exact same time the voltage drop across the resistor has to be 3V to satisfy KVL. I don't understand your statement "the potential of the charge after crossing the capacitor...will be 3V". As I already indicated, 3V is the voltage drop across the resistor when the voltage drop across the capacitor is 2 V.

Hope this helps.

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  • $\begingroup$ The potential drop is caused by the work done (per electron) in moving an electron from one plate to the other right? $\endgroup$ – Aditya Ahuja Nov 24 at 6:59
  • $\begingroup$ When you said that the remaining potential energy of the charge is given to the capacitor I picture the electron having some electric field due to the capacitor plates acting on it. The electron is able to move against the field due to the potential energy given to it by the battery . Now, since it has worked against some electric field due to the capacitor,the potential energy of the capacitor plates increases. Am I right? $\endgroup$ – Aditya Ahuja Nov 24 at 7:07
  • $\begingroup$ @Aditya Ahuja the battery gives the charge potential energy. The charge spends, drops, loses or whatever you want to call it in part as heat in the resistance, and saves the rest as electrical potential energy in the electric field of the capacitor. I’m sorry but I have no more time to spend on this. Good luck finding the answer you are looking for. $\endgroup$ – Bob D Nov 24 at 7:58

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