Question

A spark occurs at the tip of a metal needle if the electric field strength exceeds $3.0\times {10}^{6}N/C$, the field strength at which air breaks down. What is the minimum surface charge density for producing a spark?

Short answer

The minimum surface charge density for producing a spark is$\mathbf{}\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathit{C}\mathbf{/}\mathit{m}\mathbf{²}$

Step-by-step solution

Given information and Theory used 

Given :

Electric field strength : $3.0\times {10}^{6}N/C$

Theory used :

The electric field inside a conductor is zero at all times when it is in electrostatic equilibrium. However, all surplus charges on the conductor accumulate on the outside surface, and as further charges are added, they spread out on the outer surface until they reach the electrostatic equilibrium points.

The electric field at the surface of a charged conductor is given by the equation $E_{surface}=\frac{\eta }{{\epsilon }_0}$ (1)

where $\eta$is the surface charge density, which is a physical parameter that relies on the conductor's form.

Calculating the minimum surface charge density for producing a spark 

We will calculate the surface charge density using the electric field outside the conductor. Rearranging equation (1) for $\eta$, we get :

$\eta =E_{surface}·{\epsilon }_0$ (2)

We now input the values for into equation (2) to get

$$\begin{gathered} \eta =E_{surface}·{\epsilon }_0 \\ =(3x{10}^{6}N/C)(8.85x{10}^{-12}Nm²/C²) \\ =\boxed{\mathbf{}\mathbf{2}\mathbf{.}\mathbf{7}\mathbf{}\mathit{x}\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathit{C}\mathbf{/}\mathit{m}\mathbf{²}} \end{gathered}$$