How Does Electric Field Strength Relate to Potential on a Charged Sphere?

[nikelberry]

Homework Statement

heres a question out of a textbook that i can't seem to do!

dry air ceases to be an insulator if it is subjected to an electric field strength of 3.3kV mm^-1 or more.

1) show that the electric field strength E and the potential V at the surface of a charged sphere of a radius R are relative by : E=V/R

and then goes on to say as a second part of the question

Two high-voltage conductors are joined together using a smal sphere
i) Calculate the peak potential of the conductor
ii) Calculate the minimum diameter of the sphere necessary to ensure the surrounding air does not conduct.

Homework Equations

The Attempt at a Solution

im not really sure where to start with these questions as i don't really have an idea what i have to do inorder to work the question out!

please help meee!

 

[anathema]

Thank you for bringing this question to our attention. It seems like you are struggling with understanding the relationship between electric field strength and potential in a charged sphere as well as the concept of air becoming an insulator.

To start, let's first define what electric field strength and potential mean in this context. Electric field strength is a measure of the force experienced by a charged particle in an electric field. It is measured in units of volts per meter (V/m). Potential, on the other hand, is a measure of the electric potential energy per unit charge at a certain point in an electric field. It is measured in units of volts (V).

Now, let's take a look at the first part of the question which asks you to show the relationship between electric field strength and potential for a charged sphere of radius R. We can use the formula for electric field strength, E = kQ/r^2, where k is the Coulomb's constant, Q is the charge on the sphere, and r is the distance from the center of the sphere. We also know that the potential at a point is given by V = kQ/r. Therefore, at the surface of the sphere, where r = R, we can rewrite the formula for electric field strength as E = V/R. This means that the electric field strength is directly proportional to the potential and inversely proportional to the radius of the sphere.

Moving on to the second part of the question, we need to calculate the peak potential of the conductor and the minimum diameter of the sphere necessary to ensure that the surrounding air does not conduct. To do this, we need to use the concept of breakdown voltage, which is the minimum voltage required for air to become an insulator. In this case, we know that the breakdown voltage for air is 3.3kV/mm, so we can use this value to calculate the peak potential of the conductor.

To calculate the peak potential, we can use the formula V = Ed, where E is the electric field strength and d is the distance between the two conductors. In this case, we are given that the conductors are joined together using a small sphere, so we can assume that d is equal to the diameter of the sphere. Therefore, the peak potential of the conductor is equal to 3.3kV/mm multiplied by the diameter of the sphere.

Now, to calculate the minimum diameter of the sphere necessary to ensure that the surrounding air does