- #1
Electric to be
- 152
- 6
Hello. I've been doing some personal learning on the topic of dielectrics and I've come across a bit of a problem.
Say I charged up a cap. plate with a battery and then I disconnected in the battery. Then I inserted some dielectric material in the middle.
It is then given that E1 = E2 * k.
Then, since Q must remain constant, V1 / V2 = k. From this you can deduce that C1 and C2 and are a ratio of k. (Since C = Q/V)
Now I know that the capacitance will not change if there is a constant voltage applied so long as the same geometry and dielectric material applies, and therefore in the case of a constant voltage (such as putting in a dielectric material without disconnecting a battery) then k will then be a ratio of Q1 and Q2.
Now my problem:
Even though I understand this, I wanted to work backwards and prove C1 and C2 are ratios of k by showing that Q1 and Q2 are ratios of k in a constant voltage scenario. I'm having a bit of trouble doing this. I tried doing this by comparing one capacitor not connected to a battery with a dielectric plate and another capacitor connected to a battery with no dielectric plate such that both voltages (and therefore fields) are the same across each capacitor.
So then I have E1= E2, V1 = V2 and consequently (Q) = (Q1 - Qi) Well then I just get that Q = Q1 - Qi Where Q is the charge on the first capacitor plate, Q1 is the charge on the second capacitor plate, and Qi is the induced charge near the surface of the dielectric/metal plate interface. No k anywhere in sight.
How could i show that k is a ratio of Q1 and Q2 without invoking the fact that C is constant? (Like I did above to show that k is a ratio of V1 and V2)
Basically what I'm doing here is choosing to prove that C only depends on geometry and dielectric material mathematically, even though this is a known fact.
I'm probably missing something easy but would appreciate any help.
Thank you.
Say I charged up a cap. plate with a battery and then I disconnected in the battery. Then I inserted some dielectric material in the middle.
It is then given that E1 = E2 * k.
Then, since Q must remain constant, V1 / V2 = k. From this you can deduce that C1 and C2 and are a ratio of k. (Since C = Q/V)
Now I know that the capacitance will not change if there is a constant voltage applied so long as the same geometry and dielectric material applies, and therefore in the case of a constant voltage (such as putting in a dielectric material without disconnecting a battery) then k will then be a ratio of Q1 and Q2.
Now my problem:
Even though I understand this, I wanted to work backwards and prove C1 and C2 are ratios of k by showing that Q1 and Q2 are ratios of k in a constant voltage scenario. I'm having a bit of trouble doing this. I tried doing this by comparing one capacitor not connected to a battery with a dielectric plate and another capacitor connected to a battery with no dielectric plate such that both voltages (and therefore fields) are the same across each capacitor.
So then I have E1= E2, V1 = V2 and consequently (Q) = (Q1 - Qi) Well then I just get that Q = Q1 - Qi Where Q is the charge on the first capacitor plate, Q1 is the charge on the second capacitor plate, and Qi is the induced charge near the surface of the dielectric/metal plate interface. No k anywhere in sight.
How could i show that k is a ratio of Q1 and Q2 without invoking the fact that C is constant? (Like I did above to show that k is a ratio of V1 and V2)
Basically what I'm doing here is choosing to prove that C only depends on geometry and dielectric material mathematically, even though this is a known fact.
I'm probably missing something easy but would appreciate any help.
Thank you.