If a rectangular area is rotated in a uniform electric field from the position where the maximum electric flux goes through it to an orientation where only half the flux goes through it, what has been the angle of rotation?

With a uniform electric field, flux go along parallel paths, then flux is therefore proportional to the cosine of the angle rotated.
0 degree rotation => cos(0)=1 => 100% of flux.
60 degrees rotation => cos(60) => 0.5 => how many % of flux?

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ahirohit963 ahirohit963 · Answer 2 · 2021-10-05T12:00:25+02:00

Answer:

The angle of rotation is [tex]\theta=60[/tex]

Step-by-step explanation:

Given information:

The maximum flux is half to the flux after the orientation, as

[tex]\phi=\frac{\phi_m_a_x}{2}[/tex]

Find: orientation where only half the flux goes through it.

Electric Flux, number of electric field lines passing through the unit surface area.

The equation for the electric flux is,

[tex]\phi =EA\cos\theta[/tex],

Where [tex]E[/tex] is the magnitude of electric field, A is the areal vector and [tex]\theta[/tex] is the angle between the electric field direction and the areal vector.

The electric flux will be maximum when electric field direction is parallel to the surface area,

here, [tex]\theta[/tex]=[tex]0[/tex], then,

[tex]\phi_m_a_x=EA\cos{0}[/tex], [tex]\cos0=1[/tex]

[tex]\phi_m_a_x=EA[/tex],

After the orientation, the flux is,

[tex]\phi = EA{\rm{cos}\theta[/tex],

From given information,

[tex]\phi=\frac{\phi_m_a_x}{2}[/tex]

[tex]EA\cos\theta=\frac{EA}{2}[/tex]

[tex]\cos\theta=\frac{1}{2}[/tex]

[tex]\theta=60[/tex],

Hence, the angle of rotation is [tex]\theta=60[/tex]

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