# Finding the Potential Difference In Parallel Capacitors

• r0306
In summary, the conversation discusses the use of the parallel capacitor equation to find the electric fields for both plates and the total electric field between the two plates. The speaker then uses the equation E = V / d to calculate the change in V, but their answer is incorrect. The conversation suggests treating the two dielectric regions as individual capacitors and using the relationship Q = CV to calculate the potential difference across the k2 region. The speaker realizes their mistake of subtracting instead of adding the electric fields and the final answer should be 28.8*10^9 V/C.
r0306
Originally posted in a non-homework forum, so the homework template is missing
The question is based on the image below but is not the same. The question is as follows:

Assume the capacitor is charged, so that there is a charge q on the top plate and a charge -q on the bottom plate. Determine the magnitude of the potential difference across the k2 region, answering in units of q. *There is a typo in the book; k1 is the lower half of the gap. Answer in units of 109 V/C (Volt per Coulomb because this factor is multiplied by q, so the Coulomb cancels)

My attempt:

I used the parallel capacitor equation E = (Q / A) / 2(epsilon)(k) to find the electric fields for both plates. I added up the two fields and got -5.42*10^11 for the total electric field between the two plates. Then I used the equation E = V / d, plugging in my electric field and using 4.62 * 10^-3 for d and got a change in V of -1.25 * 10^9. I don't know where I went wrong can someone please help me find my mistake? Thanks.

r0306 said:
I used the parallel capacitor equation E = (Q / A) / 2(epsilon)(k) to find the electric fields for both plates. I added up the two fields and got -5.42*10^11 for the total electric field between the two plates. Then I used the equation E = V / d, plugging in my electric field and using 4.62 * 10^-3 for d and got a change in V of -1.25 * 10^9. I don't know where I went wrong can someone please help me find my mistake? Thanks.

How did you arrive at a numerical value for the field if you were not given a specific value for q?

A dielectric modifies the net electric field strength inside of it, so the field due to a given plate will be different for the two regions. You'd have to account for that. Further, they're looking for the potential across just one of the dielectric regions and you've used the entire distance between the plates, which would yield a potential difference across the the whole capacitor (if the field were in fact uniform).

An approach that you might consider is to treat the two dielectric regions as individual capacitors, where a 'virtual plate' occupies the boundary between the dielectrics. Then you can calculate the resulting capacitance of each 'capacitor' separately and apply the relationship Q = CV, where Q is the charge on the capacitor, C is the capacitance, and V the potential difference between the plates.

gneill said:
How did you arrive at a numerical value for the field if you were not given a specific value for q?

A dielectric modifies the net electric field strength inside of it, so the field due to a given plate will be different for the two regions. You'd have to account for that. Further, they're looking for the potential across just one of the dielectric regions and you've used the entire distance between the plates, which would yield a potential difference across the the whole capacitor (if the field were in fact uniform).

An approach that you might consider is to treat the two dielectric regions as individual capacitors, where a 'virtual plate' occupies the boundary between the dielectrics. Then you can calculate the resulting capacitance of each 'capacitor' separately and apply the relationship Q = CV, where Q is the charge on the capacitor, C is the capacitance, and V the potential difference between the plates.

The Q values were simply 1 and -1 respectively. Does this mean I should use half the distance given (2.31 mm) then?

EDIT: I tried using half the distance, yielding me a value of -1.25 but the answer is still wrong.

r0306 said:
The Q values were simply 1 and -1 respectively. Does this mean I should use half the distance given (2.31 mm) then?
Yes, certainly. You want to use the net field inside the k2 region and determine the potential difference across that region. The k2 region has thickness d/2.
EDIT: I tried using half the distance, yielding me a value of -1.25 but the answer is still wrong.

I got the answer. What I went wrong on was subtracting instead of adding the electric fields. The final answer should be 28.8*10^9 V/C.

## What is the formula for finding the potential difference in parallel capacitors?

The formula for finding the potential difference in parallel capacitors is V = V1 + V2 + V3 + ... + Vn, where V is the total potential difference and V1 through Vn are the individual potential differences of each capacitor.

## How do I calculate the total capacitance of parallel capacitors?

The total capacitance of parallel capacitors is calculated using the formula Ceq = C1 + C2 + C3 + ... + Cn, where Ceq is the equivalent capacitance and C1 through Cn are the individual capacitances of each capacitor.

## Can the potential difference in parallel capacitors be greater than the potential difference of any individual capacitor?

Yes, the potential difference in parallel capacitors can be greater than the potential difference of any individual capacitor. This is because the total potential difference is equal to the sum of the individual potential differences, so it can be greater than any single value.

## What happens to the potential difference in parallel capacitors if one of the capacitors is removed?

If one of the capacitors in a parallel circuit is removed, the potential difference across the remaining capacitors will stay the same. This is because the potential difference in parallel capacitors is dependent on the total charge and the total capacitance, which will not change if only one capacitor is removed.

## How does the potential difference in parallel capacitors change if the voltage source is increased?

If the voltage source in a parallel circuit is increased, the potential difference across each capacitor will also increase. This is because the potential difference is directly proportional to the voltage source, according to the formula V = Q/C. As the voltage source increases, the charge (Q) remains constant but the capacitance (C) increases, resulting in a larger potential difference.

Replies
6
Views
782
Replies
8
Views
1K
Replies
1
Views
640
Replies
7
Views
1K
Replies
18
Views
2K
Replies
3
Views
3K
Replies
30
Views
2K
Replies
11
Views
1K
Replies
14
Views
1K
Replies
7
Views
745