Question

The dielectric strength of air, E = 3.0×10⁶ V/m, is the maximum field that air can withstand before it breaks down and becomes conducting.

Part A) How much charge can be placed on a spherical conductor with a 15.0- cm radius before the field at its surface exceeds the breakdown strength of the air?

Part B) What would be the electric potential at the surface of this conductor?

Answer 1

For dielectric strength of air :

maximum electric field, E = 3 x 10⁶ V/m

radius of the sphere, r = 15 cm = 0.15 m

Part (A) : using a gauss law equation to find the value of Charge Q -

. = Q / ₀                                { eq.1 }

E A = Q / ₀       

where, A = area of the sphere = 4r²

then, E = Q / 4₀ r²

Or Q = E x 4₀ r²                                                { eq.2 }

where, ₀ = permittivity of free space = 8.85 x 10^-12 F/m

inserting the values in eq.2

Q = (3 x 10⁶ V/m) 4 (3.14) (8.85 x 10^-12 C²/N/m²) (0.15 m)²

Q = 7.5 x 10^-6 C

Part (B) : the electric potential at the surface of this conductor will be given as ::

V = k Q / r                                              { eq. 3 }

where, k = proportionally constant = 9 x 10⁹ Nm²/C²

inserting the values in eq.3,

V = (9 x 10⁹ Nm²/C²) (7.5 x 10^-6 C) / (0.15 m)

V= (67.5 x 10³ Nm²/C) / (0.15 m)

V = 450 x 10³ V

Or     V = 450 kV