Exploring the Relationship Between Parallel Capacitor Plates & Spheres

[nonvestigial]

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

 

[jbsteele]

Yes, your idea is correct. The equation C = 4πε0d is true for a spherical capacitor, which is equivalent to two parallel plates when the fringing effect is ignored. The area of the plates is related to the distance between them by A = 4πr2, where r is the radius of the sphere. This means that if you know the desired capacitance and the distance between the plates, you can calculate the area of each plate. With this information, you can then use the equation C = q/v to calculate the charge or potential difference (which are related by q = Cv).

 

[baba89]

Hello Stephanie,

Thank you for sharing your observations and thoughts on the relationship between parallel capacitor plates and spheres. I can confirm that your thinking is on the right track. The connection between the two is due to the fact that the electric field between two parallel plates is very similar to the electric field inside a spherical capacitor.

To explain this further, let's first consider the equation for capacitance, C = q/v. This equation tells us that the capacitance is equal to the charge divided by the potential difference. In the case of a parallel plate capacitor, the potential difference is equal to the electric field (E) multiplied by the distance between the plates (d). So, we can rewrite the equation as C = q/(Ed). Now, let's look at the electric field between two parallel plates. This field is constant and equal to E = V/d, where V is the voltage applied to the plates. Substituting this into our equation for capacitance, we get C = (q/d)/(V/d). The d's cancel out, and we are left with C = q/V, which is the same equation for capacitance in a spherical capacitor.

Next, let's look at the equation for potential difference, V = integration [E * ds]. This equation tells us that the potential difference is equal to the integral of the electric field over an infinitesimal distance (ds). In the case of a parallel plate capacitor, the electric field is constant, so we can pull it out of the integral. This leaves us with V = E * integration [ds]. The integration of ds is just the distance between the plates, so we can rewrite the equation as V = Ed. This is the same equation for potential difference in a spherical capacitor.

Now, let's look at your observation that setting the two capacitance equations equal to each other yields the surface area of a sphere, A = 4πr^2. This is because, in a spherical capacitor, the electric field is radial and has the same magnitude at any point on the surface of the sphere. This means that the electric field is constant and equal to E = q/(4πε0r^2), where q is the charge on the sphere and r is the radius. Substituting this into the equation for capacitance, we get C = q/(V/r) = (q * r)/V. Comparing this to the equation for capacitance in a parallel plate