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)
(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}
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.
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