Question

If the magnitude of the electric field in air exceeds roughly$\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{C}$, the air breaks down and a spark forms. For a two-disk capacitor of radius$\mathbf{50}\mathbf{}\mathbf{cm}$with a gap of role="math" localid="1656068507772" $\mathbf{1}\mathbf{}\mathbf{mm}$, what is the maximum charge (plus and minus) that can be placed on the disks without a spark forming (which would permit charge to flow from one disk to the other)? Under these conditions, what is the strength of the fringe field just outside the center of the capacitor?

Short answer

$2.1\times {10}^{-5}\mathrm{C}\mathrm{and}3000\mathrm{N}/\mathrm{C}$

Step-by-step solution

Identification of the given data

The given data is listed below as-

• The given electric field’s magnitude is,$E=3\times {10}^{-6}\mathrm{N}/\mathrm{C}$
• The radius of the capacitor is,$R=50\mathrm{cm}=50\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}=0.5\mathrm{m}$
• The gap between the capacitor i$s=1\mathrm{mm}=1\mathrm{mm}\times \frac{1\mathrm{m}}{1000\mathrm{mm}}={10}^{-3}\mathrm{m}$

Significance of the electric field, introduction of the equation of the charge for the disk and the equation of the strength for the fringe field

The electric field is described as a region that helps an electrically charged particle to exert force on another particle.

The equation of the charge for the disks is expressed as,

$E=\frac{Q/A}{{\epsilon }_0}$…(1)

Here, $Q$is the maximum charge placed on the disks, $A$is the area of the disks ($A={\mathrm{\pi R}}^2$) and ${\epsilon }_0$is the electric field constant.

The equation of the strength for the fringe field is expressed as,

$E_1=\frac{Q}{2A{\epsilon }_0}·\frac{S}{R}$…(2)

Here, $E_1$is the strength of the fringe and $S$ is the gap between the capacitors

Determination of the maximum charge that can be placed on the disk

For$E=3\times {10}^{-8}N/C$,$R=0.5m$and ${\epsilon }_0=8.85\times {10}^{-12}{C}^2/{\mathrm{Nm}}^2$in equation (1).

$$\begin{gathered} 3\times {10}^{-6}\mathrm{N}/\mathrm{C}=\frac{\mathrm{Q}}{\mathrm{\pi }{\left(0.5\mathrm{m}\right)}^2\times 8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2} \\ \mathrm{Q}=2.65\times {10}^{-15}\mathrm{C}/{\mathrm{m}}^2\times \mathrm{\pi }{\left(0.5\mathrm{m}\right)}^2 \\ =2.65\times {10}^{-15}\mathrm{C}/{\mathrm{m}}^2\times 0.785{\mathrm{m}}^2 \\ =2.09\times {10}^{-5}\mathrm{C} \\ \approx 2.1\times {10}^{-5}\mathrm{C} \end{gathered}$$

Determination of the strength of the fringe field

For $R=0.5m$,$Q=2.09\times {10}^{-5}C$ , $S={10}^{-3}\mathrm{m}$and ${\epsilon }_0=8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2$in equation (2).

$$\begin{gathered} E_1=\frac{2.1\times {10}^{-5}C}{2\times \mathrm{\pi }{\left(0.5\mathrm{m}\right)}^2\times 8.85\times {10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2}·\frac{{10}^{-3}}{0.5\mathrm{m}} \\ =\frac{2.09\times {10}^{-8}\mathrm{Cm}}{4.425\times {10}^{-12}{\mathrm{C}}^2/N}·\frac{1}{0.5\mathrm{m}} \\ =3000N/C \end{gathered}$$

Thus, the maximum charge that can be placed on the disk is$2.09\times {10}^{-5}C$ and the strength of the fringe field is$3000N/C$ .