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# A2 dielectrics help watch

1. Hi,

1. I understand from the formula E=1/2 (Q^2/C) that when the dielectric is removed the energy stored increases. The answer says it's because work is done to overcome electrostatic attraction between dielectric and adjacent plate. Does the work done contribute to the energy stored? I'm also confused as to why the energy stored increases both when you add and when you remove the dielectric.

2. I'm also really stuck on question 4 (attached) as I'm not sure how to go about starting it.

Thanks
Attached Images

2. 1. not sure what you're asking, but tbh it doesn't sound right. adding a dielectric usually increases the capacitance compared to vacuum / fresh air
http://hyperphysics.phy-astr.gsu.edu...ic/dielec.html

2. you probably want to model the capacitor as a uniform electric field... E=V/d for a uniform field
you can estimate the volume of dielectric required... that'll give you a minimum limit on the volume
3. P.S an electrolytic capacitor is a capacitor with a super dielectric allowing large capacitances in small package sizes - however they have a number of practical drawbacks, they're widely used in applications they're suitable for but can't replace all other types of capacitor.
4. (Original post by Joinedup)
1. not sure what you're asking, but tbh it doesn't sound right. adding a dielectric usually increases the capacitance compared to vacuum / fresh air
http://hyperphysics.phy-astr.gsu.edu...ic/dielec.html

2. you probably want to model the capacitor as a uniform electric field... E=V/d for a uniform field
you can estimate the volume of dielectric required... that'll give you a minimum limit on the volume
Thanks, it's a bit confusing as we are doing dielectrics before we've done electric fields.

For the question that I was trying to ask.

It asks what is the change to the energy stored by the capacitor when you add a dielectric- the answer says it increases which I understand.

The next question asks what happens to the energy stored when the dielectric is removed- the answer says it increases due to work done to overcome electrostatic forces of attraction between the dielectric and the adjacent plate- which I don't understand.
5. (Original post by VioletPhillippo)
Thanks, it's a bit confusing as we are doing dielectrics before we've done electric fields.

For the question that I was trying to ask.

It asks what is the change to the energy stored by the capacitor when you add a dielectric- the answer says it increases which I understand.

The next question asks what happens to the energy stored when the dielectric is removed- the answer says it increases due to work done to overcome electrostatic forces of attraction between the dielectric and the adjacent plate- which I don't understand.
Well you need to be quite careful about paying attention to the order in which things happen in capacitor questions (and other types of questions too - very common in capacitors)

Here are two different situations

A. A capacitor is charged to a PD of V by a constant voltage power supply and left connected to the supply, then the dielectric is removed.

B. A capacitor is charged to a PD of V and then isolated from the supply and then the dielectric is removed.

in case A because C=Q/V and V is being held constant, the decrease in capacitance must result in a reduction of the charge in the capacitor - a current will flow back into the power supply.

in case B the capacitor is isolated and the charge on it can't change... so what does C=Q/V tell us must happen if the value of C goes down?
6. (Original post by Joinedup)
Well you need to be quite careful about paying attention to the order in which things happen in capacitor questions (and other types of questions too - very common in capacitors)

Here are two different situations

A. A capacitor is charged to a PD of V by a constant voltage power supply and left connected to the supply, then the dielectric is removed.

B. A capacitor is charged to a PD of V and then isolated from the supply and then the dielectric is removed.

in case A because C=Q/V and V is being held constant, the decrease in capacitance must result in a reduction of the charge in the capacitor - a current will flow back into the power supply.

in case B the capacitor is isolated and the charge on it can't change... so what does C=Q/V tell us must happen if the value of C goes down?
Thanks so much, I understand now

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