Question

If the electric field strength in air exceeds 3.0 x 10^6 N/C, the air becomes a conductor. Using this fact, determine the maximum amount of charge that can be carried by a metal sphere 1.0 m in radius. (Hint: Review properties of conductors in electrostatic equilibrium. Also, use that the points on the surface are outside a spherically symmetric charge distribution; the total charge may be considered to be located at the center of the sphere.) Ans: ? C

Answer 1

`3.33* 10^(-4)` C

Step-by-step explanation:

`E` = Maximum electric field strength =
`3* 10^(6)` N/C

`r` = Radius of the sphere =
`1` m

`Q` = maximum charge stored by the sphere = ?

Considering that the total charge is stored at the center of the sphere, the electric field at the surface of sphere can be given as

`E=(kQ)/(r^(2))`

Inserting the values for the variables in the above equation

`3* 10^(6)=((9* 10^(9))Q)/(1^(2))`

`3* 10^(6)=(9* 10^(9))Q`

Dividing both side by
`(9* 10^(9))`

`(3* 10^(6))/(9* 10^(9))= (9* 10^(9))/(9* 10^(9))Q`

`Q = (3* 10^(6))/(9* 10^(9))`

`Q = 3.33* 10^(-4)` C