Question

If the magnitude of the electric field in air exceeds roughly$\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{N}\mathbf{3}$, the air break down and a spark forms. For a two-disk capacitor of radius 51 cm with a gap of 2 mm, if the electric field inside is just high enough that a spark occurs, what is the strength of the fringe field just outside the center of the capacitor?

Short answer

The electric field strength of fringe outside the centre of capacitor is$3\times {10}^6\mathrm{N}/\mathrm{C}$ .

Step-by-step solution

Identification of given data

The given data can be listed below,

• The magnitude of electric field of air is,$E=\frac{q}{A{\epsilon }_0}=3\times {10}^6\mathrm{N}/\mathrm{C}$
• The radius of disk is$R=51\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$
• The gap between two plates is,$r=2\mathrm{mm}\left(\frac{1\mathrm{m}}{1000\mathrm{mM}}\right)$

Concept/Significance of capacitor

the capacitor has an insulator (dielectric) inside it that prevents any charge flow through itself, so the two faces of the dielectric get charged when you charge the capacitor.

Determination of the electric field strength of fringe outside the center of capacitor

The electric field of fringe is given by.

The electric field on the capacitor is given by,

$E=\frac{q}{2A{\epsilon }_0}\frac{r}{R}$

Here, q is the charge on the disk, A is the area of disk, r is the distance between two disks and${\epsilon }_0$is the permittivity of free space.

Substitute all the values in the above equation,

$$\begin{gathered} E_{fringe}=\frac{1}{2}3\times {10}^6\mathrm{N}/\mathrm{C}\frac{2\times {10}^{-3}\mathrm{m}}{0.51\mathrm{m}} \\ =5.88\times {10}^3\mathrm{C} \end{gathered}$$

Thus, the electric field strength of fringe outside the centre of capacitor is $3\times {10}^6\mathrm{N}/\mathrm{C}$.