Consider the uniform electric field E = (2.5 j + 3.5 k) * 103 N/C. (a) Calculate the electric flux through a circular area of radius 2.5 m that lies in the yz-plane. Give your answer in N·m2/C. b) Repeat the electric flux calculation for the circular area for the case when its area vector is directed at 45° above the xy-plane. Give your answer in N·m2/C.

Given the electric field

E = (2.5• j + 3.5• k) * 10³ N/C

Given the radius of the circular path r=2.5m.

Φ=?

Flux In an electric field is given as

Φ=∮E•dA. From r=0 to r=2.5m

Given that A of a striker is

The area is in yz plane then, it's normal will be in x-direction

A=πr²

Then, dA=2πrdr •i

Φ=∮E•dA. From r=0 to r=2.5m

Φ=∮ (2.5j + 3.5k) * 10³• (2πrdr i) From r=0 to r=2.5m

Note that, i•i=j•j=k•k=1

i•j=j•i=k•i=i•k=j•k=k•j=0

Then,

Φ=10³∮ 0dr From r=0 to r=2.5m

Φ = 0 Nm²/C

b. When it is directed in the angle of 45° between above xy

Then,

dA = (2πrCos45 i + 2πrSin45 j) dr

Φ=∮E•dA. From r=0 to r=2.5m

Φ=∮ (2.5j + 3.5k) * 10³• (2πrCos45dr i + 2πrSin45 j). From r=0 to r=2.5m.

Φ=10³∮ (2.5j + 3.5k) • (2πrCos45dr i + 2πrSin45 j). From r=0 to r=2.5m.

Φ=10³∮ 5πrSin45 dr

Φ=10³*5π*Sin45 [r²/2] r=0 r=2.5m

Φ=10³ * 5π*Sin45*½[2.5²-0²]

Φ=10³ * 5π * Sin45 * ½ * 6.25

Φ=10³*34.71

Φ=34.71*10³ Nm²/C