Question

Your friend gets really excited by the idea of making a lightning rod or maybe just a sparking toy by connecting two spheres as shown in the figure below, and making R2 so small that the electric field is greater than the dielectric strength of air (3 x 10^6 N/C), just from the usual 150 V/m electric field near the surface of the Earth. If R1 is 10 cm, how small does R2 need to be, and does this seem practical? (Hint: recall the calculation for electric field at the surface of a conductor from Gauss’s Law.)

Clarification: Two isolated conducting spheres. Once connected, they must become equipotential surfaces. Use this to determine how small R2 must be in order for the electric field at the surface of the smaller sphere to exceed the dielectric strength of air.

Please show detailed work including a sketch.

R. 2 Two conducting spheres connected by a thin conducting wire.

Answer 1

Electric field at the surface of sphere, from gauss law: integral E middot ds = q/E_0 where q is charge enclosed E = q/4 pi E_0 r^2 - (1) On sphere of radius R_1 E_1 = 150 v/m (Field near earth�s surface) 1/4 pi E_0 q_1/(R_1)^2 = 150 v/m v_1 (potential) = 1/4 pi E_0 q_1/R_1 = 1/4 pi E_0 q_1/R_1 = 150 times 0.1 = 15v The spheres are at equipotential: v_1 = v_2 v_2 = 1/4 pi E_0 q_2/R_2 = 15 v - (1) E_2 = 1/4 pi E_0 q_2/R_2^2 = 3 times 10^6 N/c - (2) Thus R_2 should be given by i from equation (1) and (2) v_2/R_2 = 3 times 10^6 N/c R_2 = 15/3 times 10^6 R_2 = 5 mu m.