GenPgys2

When a piece of paper is held with one face perpendicular to a uniform electric field the flux through it is 25 N×m2/C. When the paper is turned 25° with respect to the field the flux through it is:

• 0 N×m²/C

• 11 N×m²/C

•   12 N×m²/C

•   23 N×m²/C

Answer explanation

To find the flux through the paper when it is turned at an angle of 25° with respect to the electric field, we need to use the concept of the dot product between the electric field and the normal vector of the paper.

The flux through a surface is given by the equation:

Φ = E ⋅ A

where Φ is the flux, E is the electric field vector, and A is the area vector of the surface.

When the paper is held perpendicular to the electric field, the electric field vector and the normal vector of the paper are parallel to each other, resulting in the maximum flux of 25 N·m²/C.

When the paper is turned at an angle of 25° with respect to the field, the angle between the electric field vector and the normal vector of the paper is now 90° - 25° = 65°.

The dot product of two vectors can be calculated using the equation:

A ⋅ B = |A| |B| cos(θ)

where A and B are vectors, θ is the angle between them, and |A| and |B| are their magnitudes.

In this case, the flux through the paper can be calculated as:

Φ' = E' ⋅ A

where E' is the magnitude of the electric field, and A is the magnitude of the area vector of the paper.

Since the angle between E' and A is 65°, we can rewrite the equation as:

Φ' = |E'| |A| cos(65°)

Given that the original flux was 25 N·m²/C, we can set up the following equation:

Φ' = 25 N·m²/C * cos(65°)

Using a calculator, we can evaluate the cosine of 65° and calculate the new flux Φ':

Φ' ≈ 11.38 N·m²/C

Therefore, when the paper is turned 25° with respect to the electric field, the flux through it is approximately 11.38 N·m²/C

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