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