# Discharging a charged capacitor to another neutral capacitor

• Hijaz Aslam
In summary, we are trying to find the charge on each capacitor after equilibrium is attained in the given circuit, with one capacitor initially charged to a voltage of 6.50V and the other capacitor connected in series. The method of assuming initial charges and using the equation q=CV is used to solve this problem. This method is preferred over assuming equal voltages and using the equation V1+V2=Vo, as the charge is conserved but the voltage is not in this scenario. A drawing can help clarify the concept and avoid misunderstandings.
Hijaz Aslam

## Homework Statement

After a capacitor of capacitance ##C_1## is charged up to a voltage ##V_o##, the Battery is removed from the circuit. Another capacitor of capacitance ##C_2## is connected in series with the capaciotr ##C_1##. Find the charge on each capacitor after equilibrium is attained. (##C_1=5microF##, ##C_2=5.3microF##, ##V_o=6.50V##).

##q=CV##

## The Attempt at a Solution

I tried cracking the problem as follows:

Initially ##C_1## has a voltage of ##V_o##. After connecting ##C_2## voltages of both the capacitors will become equal. So we have ##V_1=V_2## . Now as the initial potential was ##V_o##, and as the capacitors are connected in series ##V_1+V_2=V_o##. Therefore ##V_1=V_2=V_o/2##. Hence ##q_1=C_1V/2## and ##q_2=C_2V/2##. But after substituting the values, the final figure seems wrong!

My textbook attempts this question by taking: ##V_1=V_2## and assuming that ##C_1## initially had charge ##q_o##. Therefore as ##V_1=V_2## we have ##q_1/C_1=q_2/C_2## and taking ##q_1+q_2=q_o## and simplifying they arrive at the answer.
Textbook method seems fine. But what is wrong with my method? Am I missing any concepts? (I think I am likely to be wrong at ##V_1+V_2=V_o##).

Why do you think the sum of voltages stays the same?
The charge stays the same, the voltage does not have to.

Well my text mentions that in case of a Battery the voltage is maintained, whereas in case of Capacitors as it lacks the electro-chemical reaction (as in case of a cell) the voltage is not conserved. Why is the charge conserved but not the voltage. Can I have an explanation?

Why would voltage be conserved?

Dear Hijaz,
If you make a little drawing you will see that, as you already expected, ##V_1+V_2=V_o## can not be correct.
Fortunately, you have enough equations left over to determine q1 and q1

I thought about it and arrived at the following explanation : The moment ##C_2## is connected to the circuit, nature urges the voltage of both the capacitors to be the same. At the same time ##C_1## pushes charges into ##C_2## to attain the state represented by ##q_1/C_1=q_2/C_2##. And at some point when the required charge ##q_2## is consumed by ##C_2##, ##C_1## in the hurry to attain equilibrium sheds any left over voltage somehow (or any similar method).
Am I right? Or is my explanation coincidental and a blunder?

But still we can argue: why can't ##q_1/C_1=q_2/C_2## be still attained at the same time following the condition ##V_1+V_2=q_1/C_1+q_2/C_2=V_o##.
And I think I've found a possible explanation for that too. Isn't it because the charge supplied by ##C_1## is limited to ##q_o## (which ##C_1## attained during it's initial charging by the cell)?

Thinking is good. Making a little drawing is better. If you follow V1 plus V2, then you are back at the same point (*). But you traversed C2 in the opposite direction, meaning you have V1 - V2 = 0, in other words V1 = V2 -- which you already had (and correctly so).

"The moment C2 is connected to the circuit, nature urges the voltage of both the capacitors to be the same" yes ( but again: *)
"C1 pushes charge into C 2 to attain the state represented by q1 /C1 =q2 /C2" yes ( but again: *)

There is no "hurry", but it goes fast, yes. Lightning speed, so to say (there is no mention of resistance to limit the rate of discharge).
"Shedding of voltage" is the same thing as unloading charge: q = CV. Driving force is voltage difference. Unloading stops when voltage difference is zero, well before q0 is supplied.

- - - - - - -
(*)
I suddenly realize there may be a misunderstanding: how exactly do you interpret
Another capacitor of capacitance C2 is connected in series with the capacitor C1
-- again a good reason for a drawing!

## 1. How does discharging a charged capacitor to another neutral capacitor work?

When a charged capacitor is connected to a neutral capacitor, the excess charge from the charged capacitor will flow into the neutral capacitor until both capacitors have the same charge and voltage. This is known as discharging.

## 2. What happens to the charge and voltage of the two capacitors after discharging?

After discharging, both capacitors will have the same charge and voltage. This is because the charge from the charged capacitor is evenly distributed between the two capacitors, causing them to have equal amounts of charge and voltage.

## 3. Can discharging a charged capacitor to another neutral capacitor damage the capacitors?

No, discharging a charged capacitor to another neutral capacitor does not cause any damage to either capacitor. The charge simply flows from one capacitor to the other until they reach the same charge and voltage.

## 4. Is discharging a charged capacitor to another neutral capacitor reversible?

Yes, discharging a charged capacitor to another neutral capacitor is reversible. If the two capacitors are then disconnected from each other, the charged capacitor will have its original charge and voltage, and the neutral capacitor will return to its original state.

## 5. Can discharging a charged capacitor to another neutral capacitor be used for any practical applications?

Yes, discharging a charged capacitor to another neutral capacitor is commonly used in electronics for filtering and energy storage. It can also be used in circuits to protect components from high voltage spikes. Additionally, it is a crucial step in the process of charging and discharging capacitors in many electronic devices.

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