If the electric flux through a rectangular area is 5.0 Nm2/C, and the electric field is then doubled, what is the resulting flux through the area?

10.0Nm²/C

The Electric flux through an area (A) due to a electric field (E) is given by,  

ϕ  =  E A cos θ       (i)

where;

θ  is the angle between the electric field and the area.

When θ and A are held constant, equation (i) becomes;

ϕ  =  k E

ϕ / E  =  k           (ii)

Where;

k = constant =   A cos θ

Equation (ii) can thus be written as;

ϕ₁ / E ₁ =  ϕ₂ / E ₂              (iii)

Where;

ϕ₁ and E ₁ are the initial values of the flux and electric field

 ϕ₂ and E ₂ are the final values of the flux and electric field

According to the question;

E ₂ = 2 x E₁

ϕ₁ = 5.0Nm²/C

Substitute E ₂ = 2 x E₁ into equation (iii) as follows;

ϕ₁ / E ₁ =  ϕ₂ / (2 x E₁)

ϕ₁ / E ₁ =  ϕ₂ / 2E₁

ϕ₁ /  ϕ₂ = E₁ / 2E₁

ϕ₁ /  ϕ₂ = 1 / 2

ϕ₂ = 2ϕ₁

Substitute ϕ₁ = 5.0Nm²/C into the above equation as follows;

ϕ₂ = 2 x 5.0Nm²/C

ϕ₂ = 10.0Nm²/C

Therefore, the resulting flux is 10.0Nm²/C

electromagnetic shielding is the practice of reducing the electromagnetic field in a space by blocking the field with barriers made of conductive or magnetic materials. shielding is typically applied to enclosures to isolate electrical devices from their surroundings, and to cables to isolate wires from the environment through which the cable runs. electromagnetic shielding that blocks radio frequency electromagnetic radiation is also known as rf shielding.

the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia