I have a picture enclosed of the question, but I will type the question for claf

Question

I have a picture enclosed of the question, but I will type the question for claficiation: A parallel plate capacitor has square plates of side A. The plates make an angle (x) with each other. For a small x, show that; C~ ((Eo*a^2/d )*(1-ax/2d)) that is, Eo= 8.85e-12 x=theta HINT: think of a bunch of small caps in parallel this involves integration total cap = sum of cap I will only give five stars to the person who answers this thoroughly, with explanation, reasoning, and clear steps. MUST be easy to read/interpret. thank you so much :)

Explanation / Answer

consider a small cut dx

then this will have area = a* dx where a^2 = A

then distance = d + x tan theta

so dC = e0 a dx/( d + x tan theta)
since parallel we can just add these up
C = integral of dC

= integral of e0 a dx/( d + x tan theta) from 0 to a

= e0 a/d integral of dx/( 1 + x/d tan theta) from 0 to a

let u = x/d tan theta, then du = dx/d tan ttheta

and dx = d du/tan(theta)
small angle so tantheta = theta
= e0 a/d * d/theta integral of du/(1 + u) from 0 to a/d theta

= e0 a/theta * ln( 1 + u) from 0 to a/d theta

= e0 a/theta * ln ( 1 + a theta/d)

but since theta is small
ln(1 + a theta/d) = a theta/d - 1/2 (a theta/d)^2

so C = e0 a/theta (a theta/d - 1/2 (a theta/d)^2 )

c = e0 a^2/d *1 - 1/2 a theta/d)

thus shown

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