Calculating Capacitance of a Parallel Plate Capacitor Using Integration

[sankalpmittal]

Homework Statement

See the image:
http://postimg.org/image/q550yqqd1/

Left click to enlarge...

Homework Equations

For a parallel plate capacitor, Capacitance C=Kε_oA/d

A is area of plate and d is distance between the two plates..

The Attempt at a Solution

I do not know how to approach. I divided the capacitor along the diagonal into two halves. For the first half,

dC₁ = K₁ε_odA/dl

I took a strip of edge dx and length "l" perpendicular to the capacitor. Area of that strip= (x+dx)²-x²=2xdx..

dA=2xdx

Hence,
dC₁ = 2K₁ε_oxdx/dl

Now taking an angle θ, tanθ=d/a= (d-l)/(a-x)...

Am I going in the right direction ?

Please help !

Thanks in advance...

 

[gneill]

I can't clearly picture how you're dividing up the device. In my mind's eye I can see taking vertical slices of width dx across the image. At some horizontal position x, a slice will effectively consist of two capacitors in series, one with dielectric K1, the other with dielectric K2. "Plate" area is given by a*dx, and "plate" separation for each capacitor can be determined with a bit of geometry.

 

[sankalpmittal]

I can't clearly picture how you're dividing up the device. In my mind's eye I can see taking vertical slices of width dx across the image. At some horizontal position x, a slice will effectively consist of two capacitors in series, one with dielectric K1, the other with dielectric K2. "Plate" area is given by a*dx, and "plate" separation for each capacitor can be determined with a bit of geometry.

Ok, So I have a differential form,

dC₂=K₂ε_oadx/distance, as per the hint you gave. I also understand that the two capacitors are in series. So what should I replace K with ?

And about geometry... I took angle θ between diagonal and an edge. Then,

tanθ = d/a = y/x = (d-y)/(a-x)

where, the capacitor with dielectric constant K₂ is positioned at distance x and its distance from the one edge(horizontal) to the boundary of second capacitor.

Am I on the right track ? Thanks..

 

[gneill]

Ok, So I have a differential form,

dC₂=K₂ε_oadx/distance, as per the hint you gave. I also understand that the two capacitors are in series. So what should I replace K with ?
You don't replace K, rather you need to find expressions for the two capacitors in terms of their individual K's and geometry, then combine them in series to yield a net capacitance for that "slice". The slices are all in parallel...

And about geometry... I took angle θ between diagonal and an edge. Then,

tanθ = d/a = y/x = (d-y)/(a-x)

where, the capacitor with dielectric constant K₂ is positioned at distance x and its distance from the one edge(horizontal) to the boundary of second capacitor.

Am I on the right track ? Thanks..

I believe so... can you write expressions for the two capacitances of a slice?

 

[sankalpmittal]

I believe so... can you write expressions for the two capacitances of a slice?

gneill,

As per the image you gave,

dC₁ = K₁ε_oadx/d₁

dC₂= K₂ε_oadx/d₂

Correct ?

Now dC=dC₁*dC₂/(dC₂+dC₁) ?

 

[gneill]

gneill,

As per the image you gave,

dC₁ = K₁ε_oadx/d₁

dC₂= K₂ε_oadx/d₂

Correct ?

Now dC=dC₁*dC₂/(dC₂+dC₁) ?

Right. You'll want to express d1 and d2 in terms of some common variable over which you can integrate.

 

[sankalpmittal]

Right. You'll want to express d1 and d2 in terms of some common variable over which you can integrate.

Ok, so replacing d₂=d-d₁,

dC₁ = K₁ε_oadx/d₁

dC₂= K₂ε_oadx/(d-d₁)

Now, dC=dC₁*dC₂/(dC₂+dC₁)
dC=K₁K₂ε_oadx/{K₁(d-d₁)+K₂d₁}

Now to what limit shall I integrate this expression over ? Looks like I will have to eliminate d₁.

By geometry, I got,

d(a-x)/a = d₁

Correct ?

P.S. d is distance between plates as given. dC is small capacitance. dx is small distance between slices as per your image. (Thanks.)

 

[gneill]

Now to what limit shall I integrate this expression over ? Looks like I will have to eliminate d₁.

By geometry, I got,

d(a-x)/a = d₁

Correct ?

Sure. So you're integrating w.r.t. x. You should be able to see from the diagram the limits for x.

 

[sankalpmittal]

Sure. So you're integrating w.r.t. x. You should be able to see from the diagram the limits for x.

Thanks a lot gneill !

I integrated the expression in limits of x from zero to "a" and got the correct answer ! Thanks once again.