# Tag Info

8

I'm concerned with the charge of the capacitor causing the capacitor to get pulled harder by the gravity of other objects, like earth. General relativity is not required to answer this question. Consider (for simplicity) a parallel plate capacitor where the field is constrained within the parallel plates. In this case the field is uniform (let it be $E_0$) ...

7

Yes, the electric fields in a capacitor add to its weight. But not so that you’d notice with anything so crude as a balance. Suppose you had a one-farad “supercapacitor” that you could charge up to one kilovolt. The energy stored in the electric field would be $$U = \frac12 C V^2 = \frac12\times10^6 \rm\,J$$ This is an awful lot of energy for a capacitor, ...

2

There are no critical points because this function doesn't have local minima! The function is linear in $L$. To maximize, we want to make $L$ as large as we can. Assuming $b>a$, the function $\frac{1}{\ln(b/a)}$ is monotonic decreasing. To maximize, we want the ratio $b/a$ to approach $1$ from the positive direction. To summarize, if there aren't any ...

2

Calculations, based on conservation of charge, show that, for a charged capacitor, A, connected to an identical, initially uncharged, capacitor, B, the new voltage between the plates of A or B is half the original voltage between the plates of A. I assume that you can do calculations of this type, but are seeking a more intuitive reason for the voltage ...

1

You should have trouble with it. It isn't true. For example, suppose $$C_3 = 2C_1$$ $$C_2 = 10C_1$$ $$C_4 = 20 C_1$$ You will find that $$V_{ac} = V_{ad}$$ But $$Q_{right} = 10 Q_{left}$$

1

It is true, only if the capacitors have capacitance values that are symmetric ( $C_1 = C_2$ and $C_3 = C_4$ ). Else is not true. You can do the computation using the formula of a series of capacitors $1/C_{tot} = 1/C_1 + 1/C_2 +...+ 1/C_n$ (here $n = 2$). You can compute the charge by the well known relationship $Q = C \times V$.

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