# The square surface shown measures 2

The square surface shown measures 3.2 mm on each side. It is immersed in a uniform electric field with magnitude E = 1800 N/C and with field lines at an angle of θ = 35° with a normal to the surface, as shown. Take that normal to be directed “outward,” as though the surface were one face of a box. Calculate the electric flux through the surface.

Image from: J. Walker, D. Halliday, and R. Resnick, Fundamentals of physics: [extended], 10th ed. United States: Wiley, John & Sons, 2013.

#### Solution:

$\phi=(E)(\cos\theta)(A)$

(Where $E$ is electric field, $\theta$ is the angle between our area vector and electric field lines, and $A$ is area)

$\phi=(1800)(\cos180^0-35^0)(0.0032^2)$

(As the field lines are at an angle of $35^0$, our normal to surface must be at an angle of $180^0-35^0=145^0$. This calculates the angle between the electric field lines and our area vector.)

$\phi=-0.0151$$\dfrac{N\cdot m^2}{C}$

## 2 thoughts on “The square surface shown measures”

• melike ışık

An infinite line of charge produces a field of magnitude 5.4 × 104 N/C at a distance of 2.9 m. Calculate the linear charge density.

• questionsolutions Post author

You can find the linear charge density as follows:

λ=(2)(π)(ϵ)(r)(E)
λ=(2)(π)(8.85×10^(-12))(2.9)(5.4×10^4)
λ=8.7×10^(-6) C/m