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This isn't a problem out of my book, just something I noticed while *doing* a problem on parallel capacitor plates. The problem gave me an area for each plate (1 m^2)and a desired capacitance (1 F)and prompted me to find the distance between the plates.
C = q/v (capacitance = charge/potential difference)
v = integration [E*ds] (potential diff = int [electric field * infinitesimal distance])
Here's what I was supposed to be doing:
V = integration [E * ds] from 0 to d, that is Ed. From flux, q/(ε0) = EA, so q = (ε0)EA. C = q/v, therefore C = (ε0)A/d. Plug in A and C and get d.
But I was also thinking:
C = q/v. Plugging in v = kq/d, you get C=d/k or C=4π(ε0)d. Setting the two Cs equal, you get A=4πd^2, which is the surface area of a sphere! What?
I have an idea that it's related to A) an infinitesimally small piece of a sphere being a square B) ignoring fringing, and thus making the situation as though the two plates were shells. Is this right? Plugging in the d that I got (8.854 x 10^-12 m) back into A = 4πr^2 gets me a near-negligible area, which seems to make sense with the dA idea... but I'm not sure how this all plays out mathematically.
Thanks,
Stephanie
C = q/v (capacitance = charge/potential difference)
v = integration [E*ds] (potential diff = int [electric field * infinitesimal distance])
Here's what I was supposed to be doing:
V = integration [E * ds] from 0 to d, that is Ed. From flux, q/(ε0) = EA, so q = (ε0)EA. C = q/v, therefore C = (ε0)A/d. Plug in A and C and get d.
But I was also thinking:
C = q/v. Plugging in v = kq/d, you get C=d/k or C=4π(ε0)d. Setting the two Cs equal, you get A=4πd^2, which is the surface area of a sphere! What?
I have an idea that it's related to A) an infinitesimally small piece of a sphere being a square B) ignoring fringing, and thus making the situation as though the two plates were shells. Is this right? Plugging in the d that I got (8.854 x 10^-12 m) back into A = 4πr^2 gets me a near-negligible area, which seems to make sense with the dA idea... but I'm not sure how this all plays out mathematically.
Thanks,
Stephanie
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