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.

Explanation:

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

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s

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