Question

Question: Breakdown field strength for air is roughly . If the electric field is greater than this value, the air becomes a conductor. (a) There is a limit to the amount of charge that you can put on a metal sphere in air. If you slightly exceed this limit, why would breakdown occur, and why would the breakdown occur very near the surface of the sphere, rather than somewhere else? (b) How much excess charge can you put on a metal sphere of radius without causing breakdown in the neighboring air, which would discharge the sphere? (c) How much excess charge can you put on a metal sphere of onlyradius? These results hint at the reason why a highly charged piece of metal tends to spark at places where the radius of curvature is small, or at places where there are sharp points.

Short answer

Answer

(a) Electric breakdown occurs as the magnitude of the electric field is above the limit of the electric breakdown for a charged object. If the distance amongst the charged object and the point inside the electric field is decreased to a high value, then the electric field’s magnitude will exceed the value of the breakdown limit of the electric field, so that electric breakdown occurs near the sphere’s surface.

(b) The excess charge that can be put on a metal sphere is for a metal sphere of radius.

(c) The excess charge that can be put on a metal sphere is for a metal sphere of radius.

Step-by-step solution

Identification of given data

The given data is listed below as:

• The strength of the breakdown field of air is,$E=3\times {10}^6\text{N/C}$
• The radius of the metal sphere in the first case is,$R_1=10\text{cm}=10\text{cm}\times \left(\frac{\text{1}\text{m}}{\text{1} 00\text{cm}}\right)=0.1\text{m}$ .
• The radius of the metal sphere in the second case is $R_2=1\text{mm}=1\text{mm}\times \left(\frac{\text{1}\text{m}}{\text{1} 000\text{mm}}\right)=0.001\text{m}$.

Significance of the electric field

The electric field is referred to as a region that helps an electrically charged particle to exert force on another particle. The magnitude of the electric field is directly proportional to the charge induced and inversely proportional to the square of their distances.

(a) Determination of the reason for the breakdown 

Electric breakdown happens because of the reason that the electric field’s magnitude of an object that is electrically charged is mainly above the limit of the electrical breakdown of a particular insulator that mainly surrounds an object that is charged.

If the distance amongst the charged object and the point inside the electric field is decreased to a high value, then the electric field’s magnitude will exceed the value of the breakdown limit of the electric field. Hence, it is the reason the breakdown occurs near the surface of the sphere.

Thus, electric breakdown occurs as the magnitude of the electric field is above the limit of the electric breakdown for a charged object. If the distance amongst the charged object and the point inside the electric field is decreased to a high value, then the electric field’s magnitude will exceed the value of the breakdown limit of the electric field, so that electric breakdown occurs near the sphere’s surface.

(b) Determination of the excess charge for a metal sphere of radius 

The equation of the magnitude of the electric field for a metal sphere is expressed as:

$$\begin{gathered} E=k\frac{q}{{R_1}^2} \\ q=\frac{E{R}_1^2}{k} \end{gathered}$$

Here, is the magnitude of the electric field that is the strength of breakdown field of air, is the electric constant with the value , is the excess charge and is the radius of the metal sphere in the first case.

Substitute the values in the above equation.

$$\begin{gathered} q=\frac{\left(3\times {10}^6\text{N/C}\right)\times {\left(0.1\text{m}\right)}^2}{\left(9\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right)} \\ =\frac{\left(3\times {10}^6\text{N/C}\right)\times 0.01{\text{m}}^2}{\left(9\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right)} \\ =\frac{\left(30000\text{N}·{\text{m}}^2\text{/C}\right)}{\left(9\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right)} \\ =3.33\times {10}^{-6}\text{C} \end{gathered}$$

Thus, the excess charge that can be put on a metal sphere is for a metal sphere of radius.

(c) Determination of the excess charge for a metal sphere of radius

The equation of the magnitude of the electric field for a metal sphere is expressed as:

$$\begin{gathered} E=k\frac{q}{{R_2}^2} \\ q=\frac{E{R}_2^2}{k} \end{gathered}$$

Here, is the magnitude of the electric field that is the strength of breakdown field of air, is the electric constant, is the excess charge and is the radius of the metal sphere in the second case.

Substitute the values in the above equation.

$$\begin{gathered} q=\frac{\left(3\times {10}^6\text{N/C}\right)\times {\left(0.001\text{m}\right)}^2}{\left(9\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right)} \\ =\frac{\left(3\times {10}^6\text{N/C}\right)\times \left(1\times {10}^{-6}{\text{m}}^2\right)}{\left(9\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right)} \\ =\frac{\left(3\text{N}·{\text{m}}^2\text{/C}\right)}{\left(9\times {10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}\right)} \\ =3.33\times {10}^{-10}\text{C} \end{gathered}$$

Thus, the excess charge that can be put on a metal sphere is for a metal sphere of radius.