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
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