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

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.