Asked by blasting
if each parallel conducting plates have area 1, then another area 2 is added to each plates, is the charge density and charge still be the same
Answers
Answered by
bobpursley
no, the original charge disperses to reduce charge density.
Answered by
blasting
if i look at this formula Q=CV=epsilon(A1+A2)V/d, then shouldnt charge increases then charge density also increases