# Find the dimension of capacitor given [phi]E and Id

• a4pat
In summary: So, In summary, the electric field between two circular plates of a capacitor is changing at a rate of 1.5x10^6 V/m/s (ΦE). If displacement current at this instant is Id=0.80x10^-8A, find the dimensions of the plates.
a4pat

## Homework Statement

The electric field between two circular plates of a capacitor is changing at a rate of 1.5x10^6 V/m/s (ΦE). If displacement current at this instant is Id=0.80x10^-8A, find the dimensions of the plates.

## Homework Equations

Id=ΔQ/Δt=εΔΦE/Δt
Q=CV=(εA/d)(Ed)

Q=εAE <-- need to solve for A but do not have E.

## The Attempt at a Solution

Q=εΦE = (8.85x10^-12)(1.5x10^6)
Q=1.33x10^-5

Feel like I must be missing something, I've gone over this problem and relevant formulas for way too long and can't figure out how to determine A (area of the plates) from the information given. Would really appreciate a push in the right direction.
Thanks

Last edited:
What happens if you differentiate your formula Q = (εA/d)(Ed) with respect to time?

gneill said:
What happens if you differentiate your formula Q = (εA/d)(Ed) with respect to time?

Could you elaborate?

I don't see how that helps to solve for A or E

a4pat said:
Could you elaborate?

I don't see how that helps to solve for A or E

differentiate both sides w.r.t. time. (what varies on each side?). Does the result mesh with any other formula you've written?

gneill said:
differentiate both sides w.r.t. time. (what varies on each side?). Does the result mesh with any other formula you've written?

ΔQ/Δt = CV/Δt = εAE/Δt = Id

Thanks for the reply, still not sure how to solve this. Even with respect to time I can't see how the formulas can be setup to find the plate dimensions (A) without having d, or E.

a4pat said:
ΔQ/Δt = CV/Δt = εAE/Δt = Id

Thanks for the reply, still not sure how to solve this. Even with respect to time I can't see how the formulas can be setup to find the plate dimensions (A) without having d, or E.

So,

ΔQ/Δt = εA ΔE/Δt by your differentiation

and you're given:

ΔQ/Δt = 0.80x10^-8A ; and ΔE/Δt = 1.5x10^6 V/m/s

## 1. What is the relationship between [phi]E and Id in a capacitor?

The electric field strength, [phi]E, and the displacement current, Id, are directly proportional in a capacitor. This means that as one increases, the other also increases.

## 2. How do you calculate the dimension of a capacitor using [phi]E and Id?

The dimension of a capacitor can be calculated by dividing the displacement current, Id, by the electric field strength, [phi]E. This will give the unit of Farads (F), which is the unit of capacitance.

## 3. Can the dimension of a capacitor be negative?

No, the dimension of a capacitor cannot be negative. Capacitance is a physical property of a capacitor that represents its ability to store electric charge, and it cannot have a negative value.

## 4. Are there any other factors that affect the dimension of a capacitor?

Yes, the dimension of a capacitor also depends on its physical properties such as the distance between the plates, the surface area of the plates, and the type of dielectric material used.

## 5. How does the dimension of a capacitor affect its performance?

The dimension of a capacitor affects its performance by determining how much charge it can store and how quickly it can discharge. A higher dimension (or capacitance) means the capacitor can store more charge and can discharge more slowly, while a lower dimension means the opposite.

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