Calculating Energy Stored in a Parallel-Plate Capacitor

[yayovio10]

Homework Statement

A parallel-plate capacitor with plates of area (0.5m) * (1m) has a distance separation of 2 [cm] and a voltage difference of V = 200 [V], as shown in Fig.

a) Find the energy stored

b) keep d1 = 2 [cm] and the voltage difference V, while increasing d2 = 2.2 [cm]. Find the energy stored (hint hint u=1/2CV^2)

Homework Equations

The Attempt at a Solution

 

[Simon Bridge]

You could try dividing the capacitor into lots of narrow capacitors in parallel.

 

[yayovio10]

but how if d1=2 and d2=2.2, can you still use the capacitor formulas even if its a uneven capacitor?

 

[Simon Bridge]

You have to divide it into smaller capacitors than that.

 

[yayovio10]

can you give me a little example please, i just cannot imagine since one side is uneven :(

 

[Simon Bridge]

Have you not done any calculus?

 

[yayovio10]

so for example i will use this formula E = Δ V / Δ d, and by dividing the plates in very tiny plates
i can now write E as E = ∑ (Δ V / Δd), but since we V is constant E= V (∑ (1/Δd)) so this will be the same as E=V∫((1/dx)) from d1 to d2 where d=x?

is this approach correct? and then calculate the charge and then the energy?

 

[yayovio10]

ok so i was wondering if so for example i will use this formula E = Δ V / Δ d, and by dividing the plates in very tiny plates i can now write E as E = ∑ (Δ V / Δd), but since we V is constant E= V (∑ (1/Δd)) so this will be the same as E=V∫((1/dx)) from d1 to d2 where d=x?

is this approach correct? and then calculate the charge and then the energy?

 

[rude man]

Looks unwieldy. What happens to your stack of plates when you reach d = 2cm?

How about this instead: compute the E field as a function of x along the bottom plate: x = 0 on the left and x = 0.5m on the right. Compute the energy in a typical volume dx d(x) 1m and integrate.

 

[yayovio10]

the idea is that i can divide the plates in very tine ones and will use this forumula

$$Q \equiv εA(Δ V / Δd)$$ and will set $$A\equiv dxdy$$ and $$d \equiv dz$$

so the new equation will be like
$$Q \equiv εV \int_0^1 {dx} \int_0^.5 {dy} \int_a^b{1/dz}$$

where $$a\equiv2$$ and $$b\equiv2.2$$


and then use $$u \equiv QV(1/2)$$ is this correct?

 

[Simon Bridge]

Kinda... why introduce the charge?

You want to use $U=\frac{1}{2}CV^2\implies dU = \frac{1}{2}V^2\;dC$

Put the x-axis in the direction the separation of the plates varies.
Divide the width into strips L=1m long, and dx wide... this is a row of parallel plate capacitors in parallel... you know how to find the capacitance of a parallel plate capacitor.

If we put the x-axis so the separation goes like: d(x=0)=2m and d(x=0.5)=2.2m, you can find an expression for d(x)

From that, and knowing that V is the same for all the capacitor elements, you can find:
dC= (the capacitance of the element at position x)
... get this from the expression for the capacitance of a parallel plate capacitor.

Use that to get an expression of form: $dU = f(x)\;dx$, and integrate both sides.

 

[rude man]

There's no need to compute capacitance, though it can be done that way of course.

The direct way is to use the formula energy = energy density times volume. What is the energy density of a field E in vacuo?

 

[yayovio10]

using $$C\equiv \frac{εA}{d}$$ where A= area of plates and the distance is a function of x $$d(x) \equiv 2 + 0.4x$$

we get

$$C\equiv ε \frac{\int_{0}^{1}dy \int_{0}^{.5}dx}{\int_{0}^{.5}(2+.4x)dx}$$then substituting in $$U \equiv \frac{1}{2} ε V^2 C$$

we can get

$$U \equiv \frac{1}{2} ε V^2 \frac{\int_{0}^{1}dy \int_{0}^{.5}dx}{\int_{0}^{.5}(2+.4x)dx}$$

is this correct?

 

[yayovio10]

Kinda... why introduce the charge?

You want to use $U=\frac{1}{2}CV^2\implies dU = \frac{1}{2}V^2\;dC$

Put the x-axis in the direction the separation of the plates varies.
Divide the width into strips L=1m long, and dx wide... this is a row of parallel plate capacitors in parallel... you know how to find the capacitance of a parallel plate capacitor.

If we put the x-axis so the separation goes like: d(x=0)=2m and d(x=0.5)=2.2m, you can find an expression for d(x)

From that, and knowing that V is the same for all the capacitor elements, you can find:
dC= (the capacitance of the element at position x)
... get this from the expression for the capacitance of a parallel plate capacitor.

Use that to get an expression of form: $dU = f(x)\;dx$, and integrate both sides.

is this approach correct in my last reply?

 

[Simon Bridge]

I don't think so - you have too many integrals.

 

[yayovio10]

I don't think so - you have too many integrals.

where am i wrong?

 

[Simon Bridge]

Why not go step-by-step through the suggestion in post #11?

you have:
1. expression for d(x): $d=2+2x/5$ (better to avoid decimals in equations - if you can)

You still need the others:
If the plates are 1m long and dx wide - what is their area?

Therefore:
2. dC=

Therefore:
3. dU=

 

[yayovio10]

ok so after trying with my friend we got this,

is this right now?

 

[Simon Bridge]

Since you will not answer questions or follow suggestions, I cannot help you.

 

[yayovio10]

Why not go step-by-step through the suggestion in post #11?

you have:
1. expression for d(x): $d=2+2x/5$ (better to avoid decimals in equations - if you can)

You still need the others:
If the plates are 1m long and dx wide - what is their area?

Therefore:
2. dC=

Therefore:
3. dU=

is the last reply right?

 

[rude man]

OP - is your answer 4.22 μJ? Your work is very hard to read.