Answer to Question #303047 in Electricity and Magnetism for Happpy

Question

1.Consider the electric flux Φ E=E⋅A of a uniform electric field (whose magnitude is E

) that makes an angle θ with the unit vector n that is normal to a rectangular surface whose area is A. Then the electric flux ΦE can be shown to be equal to:

a. 0

b. EA

c. EA n

d. EA cos θ

2.The flux ϕE=28N⋅m^2/C through the open surface A=6i+10j+5k is given. Find the value of a if the electric field is E=−ai +8k Value of "a"

Answer 1

Answer

1.in this case flux can shown by

$\phi=EAcos\theta$

Option d is correct.

2.so flux can written

$\phi=E.A\\28=-6a+40\\a=2$