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Lets say we have a single plate that has a charge of +Q on it. A plate with charge -Q is infinite distance away. Will the plate with +Q have a capacitance associated with it? Why or why not? I was thinking that because the distance between the plates is infinity, the capacitance is zero, but according to my TA, that is not the case. There is always some sort of "absolute" capacitance for the plate with +Q (a capacitance will exist between the plate and air?). Can someone help me out in understanding this?

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  • \$\begingroup\$ Can air (or a vacuum) store/retain charge? If not then there can be no capacitance from a plate to "air" and the capacitance to the plate at infinity will be zero but, because Q=CV the voltage across the plates will be infinite for some finite charge. That's my take on it anyway (I'm better with inductors LOL) \$\endgroup\$
    – Andy aka
    Jun 1 '15 at 13:21
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$$C = \frac Q V$$

which means \$Q = C × V \$.

For a parallel plate capacitor:

$$C = \frac {8.85 × 10^{-12} \frac F m × A × k} {d} $$

Dielectric constant k = 1, since we had better assume a vacuum. Let \$A = 1 m^2\$.

$$C = \frac {8.85 × 10^{-12} Fm } {d\ m} $$

A really small number divided by a really large number.

$$\lim_{n\to \infty} \frac 1 n = 0$$

As d approaches \$\infty\$. C = 0, which means Q = 0.

There is an inverse relationship with separation, so as d gets larger, capacitance gets smaller.

So the math and I agree with you.

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You should see Spehro's answer [here], 1 If you let the distance go to infinity then the C becomes constant as \$4\pi\epsilon_0R\$. You can think about this as the energy you need to bring a test charge from infinity and stick it onto the capacitor plate.

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\$Q=CV\$

If you put some charge on the conducting plate and it has a constant potential on it which is proportional to the charge \$V=k~Q; \dfrac{1}{k}=C\$.

Like for a conducting spherical shell \$V= \dfrac{Q}{(4 \pi \epsilon R)}\$; \$C= 4 \pi \epsilon R\$.

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    \$\begingroup\$ Welcome to Electronics.SE. Your answer is hard to read due to non-existent formatting. I'm not sure what you're trying to tell us. Formula on themselves don't explain anything. \$\endgroup\$
    – Mast
    Oct 12 '18 at 6:33

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