# Kaplan FL#1: PS Q#9

Discussion in 'MCAT Study Question Q&A' started by 30somethingyearold, May 21, 2014.

1. ### 30somethingyearold

28
1
Apr 10, 2014
Hi,

I understand the answer to the qn 9 but I don't understand the reasoning for II. The question is:

The plates of the capacitor are originally separated by a vacuum. If a dielectric κ > 1 is introduced between the plates of the capacitor, and the capacitor is allowed to charge up, which of the following statements is/are true
I. The capacitance of the capacitor will increase.
II. The voltage across the capacitor plates will increase.
III. The charge stored on the capacitor will increase.

Answer = I & III (which I completely understand!). In the answer explanations is mentions that statement II is:
So statement I is correct and choice C can be ruled out. It is important to note at this point that the voltage across the plates does not increase with the introduction of a dielectric. This is because the voltage across the plates of a fully charged capacitor is equal to the voltage applied across the plates when it was initially charged. Since this is held constant at 8,000 volts, the voltage across the plates will remain at 8,000 volts. So statement II is false and choice B can be eliminated....

My thoughts:
I get that the voltage across the places don't increase with a dielectric but why does it stay the same? I thought that the voltage will decrease and capacitance will increase - what am I missing here?

3. ### tdod 5+ Year Member

218
19
Oct 30, 2011
as always, the voltage will depend on the battery (emf). the dieelectric the allows for more charge to be used to achieve the same voltage. because the emf is unchanged by the dieelectric, the voltage across the capacitor can't change.

i hope that makes sense.

4. ### 30somethingyearold

28
1
Apr 10, 2014
I'm just a bit confused about the voltage across the capacitor, I was reading a Kaplan Book online which states that when a insulating material is placed in between a charged up capacitor the voltage across it drops (lowering the voltage to leave room for charge to increase) - how is that consistent with the FL test question?