contestada

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.

Respuesta :

Kazeemsodikisola

Answer:

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

oladayoademosu

a. The electric flux through a circular area of radius 2.5 m that lies in the yz-plane is 0 Nm²/C

b. The electric flux when its area vector is directed at 45° above the xy-plane is 3.471 × 10⁴ Nm²/C

a.

The electric flux through a circular area of radius 2.5 m that lies in the yz-plane is 0 Nm²/C

Electric flux

The electric flux is given by Ф = ∫E.dA where

  • E = electric field and
  • dA = differential area vector.

Since the electric field is E = (2.5 j + 3.5 k) × 10³ N/C and the area is circular, A = πr² and dA = 2πrdr.

Since we require the flux through the yz plane, the normal to that surface is in the x - direction. So, the differential area is dA = (2πrdr)i

Electric flux through the area in the yz plane

So, Ф = ∫E.dA

Ф = ∫[(2.5 j + 3.5 k) × 10³ N/C].(2πrdr)i

Ф = ∫[(2.5 j.(2πrdr)i + 3.5 k.(2πrdr)i) × 10³ N/C]

Ф = ∫[((5πrdr)j.i + (7.0πrdr)k.i) × 10³ N/C]

Ф = ∫[((5πrdr) × 0 + (7.0πrdr) × 0) × 10³ N/C]  Since j.i = k.i = 0

Ф = ∫[(0 + 0) × 10³ N/C]

Ф = ∫[0 × 10³ N/C]

Ф = ∫0 N/C]

Ф = 0 Nm²/C

So, the electric flux through a circular area of radius 2.5 m that lies in the yz-plane is 0 Nm²/C

b.

The electric flux when its area vector is directed at 45° above the xy-plane is 3.471 × 10⁴ Nm²/C

Since the area vector is directed 45° above the xy plane, the differential area vector dA = (2πrdrcos45°)i + (2πrdrsin45°)j = (2πrdr × 1/√2)i + (2πrdr × 1/√2)j = (√2πrdr)i + (√2πrdr)j

Electric flux through area directed 45° to the xy plane

So, the electric flux, Ф = ∫E.dA

Ф = ∫[(2.5 j + 3.5 k) × 10³ N/C].[(√2πrdr)i + (√2πrdr)j]

Ф = ∫[(2.5 j.(√2πrdr)i + 3.5 k.(√2πrdr)i + 2.5j.(√2πrdr)j + 3.5k.(√2πrdr)j) × 10³ N/C]

Ф = ∫[(2.5(√2πrdr)j.i + 3.5(√2πrdr)k.i + 2.5(√2πrdr)j.j + 3.5(√2πrdr)k.j) × 10³ N/C]

Ф = ∫[(2.5(√2πrdr) × 0 + 3.5(√2πrdr) × 0 + 2.5(√2πrdr) × 1 + 3.5(√2πrdr) × 0) × 10³ N/C] Since j.i = k.i = k.j = 0 and j.j = 1

Ф = ∫[(0 + 0 + 2.5(√2πrdr) + 0) × 10³ N/C]

Ф = ∫[(2.5(√2πrdr)) × 10³ N/C]

We integrate r from r = 0 to 2.5 m since r = 2.5 m.

So, Ф = ∫[(2.5(√2πrdr)) × 10³ N/C]

Ф = (2.5(√2π)∫rdr) × 10³ N/C

Ф = (2.5(√2π)[r²/2]₀²) × 10³ N/C

Ф = (2.5(√2π)[(2.5 m)² - (0 m)²]/2) × 10³ N/C

Ф = (2.5(√2π)[(6.25 m²) - (0 m²)]/2) × 10³ N/C

Ф = (2.5(√2π)[6.25 m²]/2) × 10³ N/C

Ф = (2.5(√2π)[3.125 m²]) × 10³ N/C

Ф = (7.8125√2π m²]) × 10³ N/C

Ф = (24.544√2 m²]) × 10³ N/C

Ф = 34.71 m²] × 10³ N/C

Ф = 3.471 × 10⁴ Nm²/C

The electric flux when its area vector is directed at 45° above the xy-plane is 3.471 × 10⁴ Nm²/C

Learn more about electric flux here:

https://brainly.com/question/5932170

Otras preguntas