Mythological thunderbolts

For this problem we can model Zeus' hands as a parallel plate capacitor where the plates are 'very close together'.

The Capacitance of a capacitor is defined as the charge per unit voltage. $C = \frac{q}{V}$ . This can be rearranged to give $q = CV$ . If we can find the capacitance and potential difference across this capacitor then we can find the charge induced.

To find the capacitance, we can use the approximation $C \approx \epsilon_r \epsilon_0 \frac{A}{d}$ where C is capacitance; A is the area of the plates; d is the distance between the plates. $\epsilon_0$ is the constant permittivity of a vacuum and $\epsilon_r$ is the relative static permittivity. Here, $\epsilon_r$ is just one. Note that this approximation only works because the field is assumed to be uniform since Zeus' hands are close together.

Converting the length and width of Zeus' hands from centimeters to meters and plugging the numbers in, we obtain

$C = 1 \times 8.854 \times 10^{-12} \times \frac{0.15 \times 0.2}{d}$

$C = \frac{2.6562 × 10^{-13}}{d}$

Since we do not know the distance between the plates we can not obtain a numerical value for capacitance. As we shall see, this problem resolves itself.

Next we find the voltage required to ionize air and hence create lightning. The Electric field strength required is given in the question in volts per metre and if we once again assume a uniform field, the voltage is given by $E = V/d$ or $V = Ed$

$V = 3 \times 10^6 d$

Returning to our original equation:

$q = CV$

$q = \frac{2.6562 × 10^{-13}}{d} \times 3 \times 10^6 d$

The distance cancels to give the required answer.

Zeus' hands can be modelled as two parallel conductive plates separated by a dielectric (air) with permittivity $ε_0$

Dielectric permittivity $ε_0 = 8.854187817 \times 10^{-12}$ farads per meter

Area of the plates= 0.2m $\times$ 0.15m = 0.03 $m^2$

The threshold for air ionisation is $E = 3 \times 10^6 V/m$

E = Electric field strength V = Potential difference d = Distance between the plates Q = Charge $ε_0$ = Dielectric permittivity A = Area of the plates

E = V/d

Q = C.V

C = $ε_0$ .A/d

Therefore

Q = $ε_0$ *A/d * E * d

Q = $ε_0$ * A * E

Q = $8.854187817 \times 10^{-12} * 0.03 * 3 \times 10^6$

Q = 797 nC

If the distance between Zeus's hands is much smaller than the hands themselves, then we effectively have a capacitor. Assuming that the charges are uniformly distributed over the hands / plates, the electric field between the plates is constant, with a magnitude of

$|E| = \frac{\sigma}{\epsilon}$ ,

where $\sigma$ and $-\sigma$ are the surface charge densities of the plates and $\epsilon$ is the permittivity of the air. The surface charge density is simply

$\sigma = \frac{\rho}{A}$ ,

where $\rho$ is the total charge and $A$ is the area of each plate. For the given required value of $E$ , we can thus work out what $\rho$ needs to be:

$\rho = \sigma A = \epsilon E A$ .

With $\epsilon \approx \epsilon_0 = 8.85 \cdot 10^{-12}\ \mathrm{C/(Vm)}$ , $E = 3 \cdot 10^6\ \mathrm{V/m}$ and $A = 20\ \mathrm{cm} \times 15\ \mathrm{cm} = 0.03\ \mathrm{m^2}$ , we compute

$\rho = 7.97 \cdot 10^{-7}\ \mathrm{C} = 797\ \mathrm{nC}$ .

Treat Zeus's palms as a parallel plate capacitor. Let $c$ be the capacitance, potential difference across the plates to be $v$ Volts. Distance between the plates be $d$ and area of the plates be $A$ . From the given values of length and breadth of Zeus's palms we find area $A$ to be $0.03$ . We have the equations $c = \frac{\epsilon_{o}A}{d}$ and $c = \frac{q}{v}$ where $q$ is the charge to be induced on the plates. Rearranging for $q$ and plugging in given values and taking $\frac{v}{d}$ as $E = 3 \times 10^{6} V/m$ we get $q = 796.5 nC$ .

The problem can be solved by considering the hands of Zeus as the plates of a parallel plate capacitor and then using the following formula which basically defines the Capacitance of a parallel plate capacitor :

$\frac {\epsilon \times a}{d} = \frac {q}{v}$

Here $\epsilon$ is the permittivity of medium, $a$ is the area of the parallel plates, $d$ is the distance between the plates, $q$ is the charge on each plate and $v$ is the potential difference between the plates.

Now, considering the situation, $\epsilon$ is the permittivity of free space ( around $8.854 \times 10^{-12}$ , in SI units ) and the area ( $a$ ) is $20 \times 15 \times 10^{-4}$ , in SI units. The electric field required to break down air is given in question to be $3 \times 10^6$ , in SI units. This electric field will be generated due to the potential difference working between those plates and thus, $E = \frac {v}{d}$ . Solving for $q$ , we get the final answer.

If Zeus induces equal but opposite charge on his parallel palms, he essentially makes a parallel plate capacitor. Once the electric field inside this capacitor reaches the threshold value, air becomes ionized and sparks occur which lead to thunderbolts.

Electric field inside the parallel plate capacitor is given by $E = \frac{\sigma}{\epsilon_0} = \frac{q}{A\epsilon_0}$ . Since the palms are rectangular, $A = a b = 0.03~\mbox{m}^2$ . Therefore, $q = A \epsilon_0 E = 796.5~\mbox{nC}$ .