A 0.1 m by 0.1 m sheet of cardboard is placed in a uniform electric field of 10 N/C. At first, the plane of the sheet is oriented perpendicular to the electric field vector so that the electric flux through the sheet is 0.01 N-m2/C. By what angle do you need to rotate the sheet to reduce the electric flux by 1/2?

Respuesta :

Answer:

The angle is 89°.

Explanation:

Given that,

Electric field = 10 N/C

Electric flux = 0.01 N-m²/C

Area [tex]A=\pi\times(0.1)^2[/tex]

We need to calculate the angle

Using formula of electric flux

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

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

Where, E = electric field

[tex]\phi[/tex] = electric flux

A = area

Put the value into the formula

[tex] \cos\theta=\dfrac{\dfrac{0.01}{2}}{10\times\pi\times(0.1)^2}[/tex]

[tex]\theta=\cos^{-1}(0.01592)[/tex]

[tex]\theta=89.0^{\circ}[/tex]

Hence, The angle is 89°.

The angle required to rotate the sheet to reduce the electric flux by 1/2 is 89 degrees.

What is electric flux?

The number of electric lines that interact the area of a given object or space.

It can be given as,

[tex]\phi=ES\cos \theta[/tex]

Here, [tex]E[/tex] is the magnitude of electric field, [tex]S[/tex] is the area of surface and [tex]\theta[/tex] is the angle between electric field and perpendicular.

Given information-

The dimensions of sheet of cardboard is 0.1 by 0.1.

The sheet is placed between the uniform electricity field of 10 N/C.

Put the values in the above formula as,

[tex]\dfrac{0.01}{2} =10\times\pi\times{0.1^2}\times\cos \theta\\\theta=cos^-(0.01592)\\\theta=89^o[/tex]

Hence the angle required to rotate the sheet to reduce the electric flux by 1/2 is 89 degrees.

Learn more about the electric flux here;

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