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 10⁶ 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.

spheres as shown in the figure below, and making R, so small that the electric field is greater than the dielectric strength of air (3 x 10 N/C), just from the usual 150 V/m electric field near the surface of the Earth. If R, is 10 cm, how small does R, 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 e .) 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. Refer to Chapter 7 in the openstax text for more help 2 Two conducting spheres connected by a thin conducting wire < Previous

please give a step by step solution

Answer 1

Electric field at the surface of sphere, from Gauss aw: integral E . ds = q/epsilon_0 where q is charge enclosed. E = q/4 pi epsilon_0 r^2 - 1 On sphere of radius R_1 E_1 = 150 V/m (field near earth�s surface) 1/4 pi epsilon_0 q_1/(R_1)^2 = 150 v/m v_1 (potential) = 1/4 pi epsilon_0 q_1/R_1 = 150 times 0.1 = 15 v \r\n since sphere�s are at equipotential v_1 = v_2 v_2 = 1/4 pi epsilon_0 q_2/R_2 = 15 v - 1 E_2 = 1/4 pi epsilon_0 q_2/R^2_2 = 3 times 10^6 N/C - 2 [Dielectric strength Thus R_2 should be given by from eq 1 and 2 v_2/R_2 = 3 times 10^0 N/c R_2 = 15/3 times 10^6 R_2 = 5 mu m