Figuring out how to approach capacitor problem

In summary, the conversation discusses the concept of capacitors in a circuit and how they are affected by the insertion of a dielectric. It is established that, in a series circuit, the charges on the two capacitors are always equal and the voltage drop across each capacitor is directly proportional to its capacitance. With the insertion of a dielectric, the capacitance of one capacitor increases while the other remains the same, resulting in a change in the voltage drops across each capacitor. However, the overall charge on each capacitor remains the same.
  • #1
RoboNerd
410
11

Homework Statement



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Homework Equations



C = q / voltage

The Attempt at a Solution



OK.
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By Kirchoff's Voltage Law, we have:

EMF - ( q / Capacitance)x - (q/Capacitance)y = 0

Now. I know that having the dielectric increases the capacitance of X, so the voltage drop across it is smaller, which is in accordance with D being the right answer.

On the other side, I thought that the Charge Qy would have to increase to create a higher voltage drop across capacitor Y to compensate for the decreased voltage drop across capacitor X.

However, the right answer says that the charges on the two capacitors are the same. Why is this the case?

Thanks in advance for the help!
 
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  • #2
That is how I look at it

Imagine a circuit with just wires with resistance, The current is the same through any point in the circuit because if that wasn't the case we would either see charges disappear(wrong) or see charges build up (Which I think they do for a matter of a really short time until it gets to steady state) but in the steady state we don't see that which means the current through the circuit is the same

This applies to capacitors too, So the current going in capacitor X is the same as the current going in capacitor Y and by I = Q t they have the same charge :D

Or you can even imagine the current as you can't be faster than the slowest current.
 
  • #3
RoboNerd said:
Now. I know that having the dielectric increases the capacitance of X, so the voltage drop across it is smaller, which is in accordance with D being the right answer.
The voltage across capacitor X is smaller than across capacitor Y.

RoboNerd said:
On the other side, I thought that the Charge Qy would have to increase to create a higher voltage drop across capacitor Y to compensate for the decreased voltage drop across capacitor X.

The charge is the same on both capacitors, as they are connected in series, but you are right, this charge increases with respect to that before inserting the dielectric.
In case of a voltage source of emf E and capacitors C, what is the charge and voltages Vx and Vy before and after inserting the dielectric?
 
  • #4
RoboNerd said:
On the other side, I thought that the Charge Qy would have to increase to create a higher voltage drop across capacitor Y to compensate for the decreased voltage drop across capacitor X.
Just pointing out a thing, First, not only Qy that increases both Qy and Qx increase and they are always equal.
Second, To compensate the decreased voltage not only the Cy is responsible for increasing the voltage but both of them a Q charge gets added to both of them until their combined voltage reach the source voltage
 
  • #5
RoboNerd said:
On the other side, I thought that the Charge Qy would have to increase to create a higher voltage drop across capacitor Y to compensate for the decreased voltage drop across capacitor X.

However, the right answer says that the charges on the two capacitors are the same. Why is this the case?

Thanks in advance for the help!

The misconception that it is anything else for capacitors in series seems to be quite common among students. I have been explaining the point in a number of answers, e.g. the link below. Just keep hold of the fact that you always have the overall neutrality really. That the 'inside' pair of plates |-| cannot be anything but neutral overall, because they are not conductively connected to any source of electricity! There is just a charge separation the + on one plate equalling the - on the other. Think, it cannot be otherwise. (So that reduced your choices of answers to two.) And the capacitors themselves -||- it is strictly a misleading manner of speaking to say as you constantly hear that they 'store charge'; overall they are neutral the + on one plate equalling the - on the other, so what they really 'store' is a very local separation of charges.

#4
 
  • #6
Thanks everyone for the help! Gladly appreciated.

Biker said:
Qy that increases both Qy and Qx increase and they are always equal.
I forgot the fact that the charges have to be equal for two capacitors in series. Thanks for catching that!

ehild said:
In case of a voltage source of emf E and capacitors C, what is the charge and voltages Vx and Vy before and after inserting the dielectric?

For before a dielectric, I use KVL and I have: EMF - (Q / C)x - (Q/C)y = 0 -------> Voltage drop is thus EMF/2 per each capacitor and the charge is going to be equal to EMF* Capacitance /2

After adding a dielectric, I use KVL and I have: EMF - (q / (5C) ) - (q/ C) = 0 -> Charge on each capacitor is going to be equal to (5 / 6) * (EMF * Coriginal). Thus, the voltage drop across Cap. X is going to be the charge divided by the newercapacitance capacitance: (EMF)/6 and the voltage drop across Cap. Y is going to be (5/6)*(EMF/6).

Right? Thanks for the help!
 
  • #7
RoboNerd said:
Thanks everyone for the help! Gladly appreciated.I forgot the fact that the charges have to be equal for two capacitors in series. Thanks for catching that!
For before a dielectric, I use KVL and I have: EMF - (Q / C)x - (Q/C)y = 0 -------> Voltage drop is thus EMF/2 per each capacitor and the charge is going to be equal to EMF* Capacitance /2

After adding a dielectric, I use KVL and I have: EMF - (q / (5C) ) - (q/ C) = 0 -> Charge on each capacitor is going to be equal to (5 / 6) * (EMF * Coriginal). Thus, the voltage drop across Cap. X is going to be the charge divided by the newercapacitance capacitance: (EMF)/6 and the voltage drop across Cap. Y is going to be (5/6)*(EMF/6).

Right? Thanks for the help!

Vy is wrong, there are too many "6"-s. :smile:
 
  • #8
Sorry I typed a bit extra. But apart from that?
 
  • #9
RoboNerd said:
Sorry I typed a bit extra. But apart from that?
Apart from that, it is correct. :smile:
 
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Likes RoboNerd
  • #10
Thanks for the help!
 

1. How do capacitors work?

Capacitors are electronic components that store electrical energy in an electric field between two conductive plates. When a voltage is applied, one plate accumulates a positive charge and the other plate accumulates a negative charge, creating a potential difference between the plates. This potential difference allows the capacitor to store energy, which can be released when needed.

2. What is the equation for calculating the capacitance of a capacitor?

The capacitance of a capacitor is determined by the equation C = Q/V, where C is the capacitance in Farads, Q is the charge stored on the capacitor in Coulombs, and V is the voltage across the capacitor in Volts.

3. How do I determine the charge or voltage of a capacitor given its capacitance?

To determine the charge on a capacitor, use the equation Q = CV, where Q is the charge in Coulombs, C is the capacitance in Farads, and V is the voltage across the capacitor in Volts. To determine the voltage across a capacitor, use the equation V = Q/C, where V is the voltage in Volts, Q is the charge in Coulombs, and C is the capacitance in Farads.

4. What is the time constant of a capacitor?

The time constant of a capacitor is the amount of time it takes for the capacitor to charge or discharge to 63.2% of its maximum charge or voltage. It is determined by the equation τ = RC, where τ is the time constant in seconds, R is the resistance in Ohms, and C is the capacitance in Farads.

5. How do I solve complex capacitor problems with multiple capacitors?

To solve complex capacitor problems with multiple capacitors, you can use the principles of series and parallel capacitors. In series, the total capacitance is equal to the reciprocal of the sum of the reciprocals of the individual capacitances. In parallel, the total capacitance is equal to the sum of the individual capacitances. You can use these principles, along with the equations mentioned above, to solve for various parameters such as charge, voltage, and capacitance in a circuit with multiple capacitors.

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