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Question: An electric field given by E = 4\(\widehat{i}\) – 3(y<sup>2</sup> + 2) \(\widehat{j}\) pierces Gauss...

An electric field given by E = 4i^\widehat{i} – 3(y2 + 2) j^\widehat{j} pierces Gaussian cube of side 1m placed at origin such that its three sides represent x, y and z axes. The net charge enclosed within cube is-

A

0

B

0

C

0

D

Zero

Answer

0

Explanation

Solution

E1 = 3(12 + 2) = 9

E2 = 3(0 + 2) = 6

Net flux = (9 – 6) × |x|

qϵ0\frac { \mathrm { q } } { \epsilon _ { 0 } }= 3

q = 3Ī0