IPhO 1993, Теория — Задача 1. Atmospheric Electricity

1) Surface charge

By Gauss' law, $\sigma=\varepsilon_0E_0$.

The source's intermediate product prints positive $1.3\cdot10^{-9}$ and a negative final charge; this printed sign discrepancy is retained.

2) Atmospheric charge

Consider a cylinder of cross-section $A$ with faces at heights of 0 and 100 m.

By Gauss' law,

3) Neutralization and alternative estimate

If a conductor contains $n$ charges per unit volume, each with charge $q$ and travelling with speed $v$, the current per unit area ($j$) is given by:

Here, we have both positive and negative charges ($\pm e$). Clearly, with a downward electric field, the positive charges move downward and the negative charges move upward. In the situation as described, only the positive charges can contribute to neutralization of the Earth's surface charge. Hence we have (taking downward as the positive direction for this purpose):

Now $j$ is the rate of change ($d\sigma/dt$) of the surface charge density, $\sigma$, and $E$ (if defined as positive downward) is equal to $-\sigma/\varepsilon_0$. Thus the above equation can be written:

This is just like the equation of radioactive decay. Its solution is an exponential decrease of $\sigma$ with time:

Putting $\sigma(t)=\sigma_0/2$ then gives $t=\tau\ln2=0.693\times600\approx415\ \mathrm{sec}\approx7\ \mathrm{min}$.

[A simpler approximate solution is to assume that $j$ remains constant at its initial value $j_0$:

With $|\sigma_0|=1.3\times10^{-9}\ \mathrm{C/m^2}$ from part 1, we would then put:

]

4) Field-mill induced charge

If $t=0$ is an instant at which the insulated quadrants are completely shielded, we have the following relations:

Corresponding variations occur during all the succeeding pairs of quarter-cycles.

The maximum (negative) induced charge is given by:

5) Amplifier response in both limits

This question can be discussed without making a full circuit analysis. One only needs to realize that the rate of flow of charge into the amplifier is divided into a rate of charging of the capacitor, $C\,dV/dt$, and a conduction current, $V/R$, through the resistor. There are then two extreme situations, depending on whether the amount of charge lost by leakage during one quarter-period is small or large compared to $CV$.

(a) If $CV\gg(V/R)\times(T/4)$ — i.e., $T=T_a\ll CR$ — very little of the charge is carried away through $R$ during the time $T/4$. Thus, when the insulated quadrants are charged negatively through induction, an almost equal positive charge is given to $C$. Thus $V(t)$ rises almost linearly with $t$ between $t=0$ and $t=T/4$, and then decreases almost linearly by an equal amount between $t=T/4$ and $t=T/2$. In this case,

where $q_{\max.}$ has the value obtained in part 4.

Note: Ultimately (unless $CR$ is infinite) the form of $V_a$ will become a sawtooth varying symmetrically between $\pm q_{\max.}/2C$. The statement of the problem avoids this complication by specifying that $V$ is measured just after the rotation has begun.

(b) If, however, $T=T_b\gg CR$ — i.e., $CR\ll T_b$ — most of the charge is quickly carried away through $R$. A constant positive current flows through $R$ when the magnitude of $q$ is increasing, and an equal negative current when the magnitude of $q$ is decreasing. The size of this current is approximately equal to $|q_{\max.}|/(T_b/4)$. The resulting voltage across $R$ is approximately constant during each quarter-period, and is alternately positive and negative. In this case,

The last native expression uses the signed $q_{\max.}$, while its preceding current expression uses $|q_{\max.}|$; both are preserved.

Putting these results together, we see that:

6) Numerical maximum voltage

We have $CR=10^{-8}\times2\cdot10^7=0.2\ \mathrm s$, and $T=1/50=0.02\ \mathrm s$.

Thus $CR=10\times T$, which satisfies the criterion $CR\gg T$. Therefore we can use the solution 5(a) above.

We have

and so

This numerical part uses positive charge magnitudes as printed, following the negative signed charges in parts 1 and 4.

Grading Scheme

Theoretical Problem 1: Grading Scheme