Berkeley Review--electric fields

[Mr Brain]

Question 55 from p. 146 asks

A capacitor that contains a dielectric is fully charged by and subsequently disconnected from a battery. removing the dielectric form this charged capacitor decreases its capacitance because:

A. the voltage across the plates decreases
B. the voltage across the plates increases
C. the amount of charge on each plate decreases
D. the amount of charge on each plate increases

I used the formulas and chose B, which is the correct answer. but could someone please explain the above in a conceptual manner? The capacitor is fully charged initially, meaning that the max amount of charges are put on each plate of the capacitor. The dielectric restricts the flow of charge (i.e. current) before the battery is disconnected. Once the battery is disconnected and the dielectric removed, charge is able to flow from one plate to another. Capacitance, or ability of the plates to hold charge, decreases... so why would voltage across the plates increase? Shouldn't the voltage/potential difference across the plates decrease? (I know the formula C = q/V says voltage should increase, but I still don't understand...)

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[Postictal Raiden]

Question 55 from p. 146 asks

A capacitor that contains a dielectric is fully charged by and subsequently disconnected from a battery. removing the dielectric form this charged capacitor decreases its capacitance because:

A. the voltage across the plates decreases
B. the voltage across the plates increases
C. the amount of charge on each plate decreases
D. the amount of charge on each plate increases

I used the formulas and chose B, which is the correct answer. but could someone please explain the above in a conceptual manner? The capacitor is fully charged initially, meaning that the max amount of charges are put on each plate of the capacitor. The dielectric restricts the flow of charge (i.e. current) before the battery is disconnected. Once the battery is disconnected and the dielectric removed, charge is able to flow from one plate to another. Capacitance, or ability of the plates to hold charge, decreases... so why would voltage across the plates increase? Shouldn't the voltage/potential difference across the plates decrease? (I know the formula C = q/V says voltage should increase, but I still don't understand...)

I will try to answer this but I am not sure if I am using the correct approach. I think that the key word in this question is "disconnected"; which means that the charge will remain constant. From the given equation, if u know that the capasitance is decreasing and the charge remains the same, u could conclude that voltage, V, has to increase.

again, someone please corrects me if I am wrong.

 

[Fort]

The charge on the plates of the capacitor has no where to go, so it stays stuck to the plates. The issue is the electric field between the plates. With the dielectric in place, the battery will have to force more charge onto the plates of the capacitor to maintain the same voltage.

Think of the dielectric as something that dampens an existing electric field. Since the electric field is weaker, the battery will pump more charges onto the plates of the capacitor to maintain the same field strength between the plates and thus maintain the same voltage. Now when you pull the dielectric out, the electric field actually increases and will be stronger than if you had just charged the capacitor without the dielectric. Now that the electric field is stronger, the voltage across the plates must have increased as well.