Question

Prove the following theorem: If a number of conducting surfaces are fixed in po- sition with a given total charge on each, the introduction of an uncharged, insulated conductor into the region bounded by the surfaces lowers the electrostatic energy,

Answer 1

n conducting Surfaces with bounding surfaces Si and total charge Zi (2=1,2,...,n) . The electric field in this case is Ē'. the electrostatic energy W, 'The volume exterior to si we designate V. Now introduce another conductor with bounding Surface S. and volume vo We designate volume exterior to both Si & So, V = V-Vo The electric field is now E' and the electrostatic energy w. Now electrostatic energy difference can be written a w-wie SEdie-Sveide - SE? de +5,C6-E") ? - the SE2dx + SCEvde "+ GECE E')de] The last term in the above v expression will be zero when we deal with equipotential Surfaces, Reason :- 'CÈ-' .-74.. (- E') *-COLACE ] - F. CĒS ES) Since F.(-E') :0 we have E CE E ") = -0 Fi CEE") so, SELLE E) A = -5°27 Esb. f da sn

So, we end up with W-w'= o CE?dze + S S CĒ - E') dx > 0 so, wyw' hence electrostatic energy is lowered.