Figuring out how to approach capacitor problem

[RoboNerd]

Homework Statement

Homework Equations

C = q / voltage

The Attempt at a Solution

OK.
[/B]
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!

 

[Biker]

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.

 

[ehild]

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.

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?

 

[Biker]

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

 

[epenguin]

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

 

[RoboNerd]

Thanks everyone for the help! Gladly appreciated.

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!

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!

 

[ehild]

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.

 

[RoboNerd]

Sorry I typed a bit extra. But apart from that?

 

[ehild]

Sorry I typed a bit extra. But apart from that?

Apart from that, it is correct.

 

[RoboNerd]

Thanks for the help!