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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

Respuesta :

Answer:

[tex]3.33\times 10^{-4}[/tex] C

Explanation:

[tex]E[/tex] = Maximum electric field strength = [tex]3\times 10^{6}[/tex] N/C

[tex]r[/tex] = Radius of the sphere  = [tex]1 [/tex] m

[tex]Q[/tex] = 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

[tex]E=\frac{kQ}{r^{2}}[/tex]

Inserting the values for the variables in the above equation

[tex]3\times 10^{6}=\frac{(9\times 10^{9})Q}{1^{2}}[/tex]

[tex]3\times 10^{6}=(9\times 10^{9})Q[/tex]

Dividing both side by [tex](9\times 10^{9})[/tex]

[tex]\frac{3\times 10^{6}}{9\times 10^{9}}= \frac{9\times 10^{9}}{9\times 10^{9}}Q[/tex]

[tex]Q = \frac{3\times 10^{6}}{9\times 10^{9}}[/tex]

[tex]Q = 3.33\times 10^{-4}[/tex] C

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