Question

prove that : in gasuas law for electric flux . EA=Q in /abslon nod = density fo infinte charged plane /2 abslon nod

Answer 1

\[\mathbb{Q} \div \epsilon = EA = \sigma \div 2\epsilon ... \in infinte planes \]

Answer 2

The second equality is dimensionally incorrect.

Answer 3

it was proven on a cylinder :S

Answer 4

\(\Large \frac {\sigma}{2\epsilon_o}\) is already an electric field. It CANNOT be an electric field multiplied by an area.

Answer 5

right i want to find electric flux which equal electic field *area

Answer 6

can you till me the constant ϵo equals to what ?

Answer 7

8.851*1069 ?

Answer 8

*10^9 *** ??

Answer 9

\(\large \epsilon_o=\LARGE \frac {1} {4\pi10^{-7} c^2} \large\approx 8{,}85\times 10^{-12}F.m^{-1}\)

Answer 10

snx :)

Answer 11

yw :))