Answer to Question #168887 in Electric Circuits for Marbie gonzales

Question

You are holding a 35 cm x 42 cm sheet of paper horizontally when a thundercloud produced a vertical electric field of magnitude of 45 kN/C at ground level(a) What is the electric flux through the sheet? (b) What would the flux be if you tilt the sheet of paper by 55°? (c) What would the flux be if you hold the sheet of paper vertically?

Answer 1

Explanations & Calculations

• Flux is how many field lines go through a unit area perpendicular to it & it is given in the equation form as

$\qquad\qquad \begin{aligned} \small \phi &= \small A\cdot\vec{E}\sin\theta \end{aligned}$ where $\small \theta$ is the angle by how much the considered plane is tilted with the electric field.

• Using this, those needed answers can be found

• Area of the sheet

$\qquad\qquad \begin{aligned} \small A&= \small 0.35m\times0.42m=0.147m^2 \end{aligned}$

a)

• As the sheet is horizontal, it is perpendicular to the field, then,

$\qquad\qquad \begin{aligned} \small \phi_1&= \small 45\times 10^3NC^{-1}\times 0.147m^2\\ &= \small 6615\,Nm^2C^{-1} \end{aligned}$

b)

• As the plane is tilted it is $\small (90^0-55^0=)35^0$ slanted from the verticle: hence from the field. Then,

$\qquad\qquad \begin{aligned} \small \phi_2 &= \small 45\times 10^3\times 0.147m^2\times \sin35\\ &= \small 3794.21\,Nm^2C^{-1} \end{aligned}$

c)

• As the plane is kept verticle in this orientation, it makes $\small 0^0$ with the field, which means no field line passes through the plane.
• Hence the flux is zero

$\qquad\qquad \begin{aligned} \small \phi_3&= \small 45\times 10^3\times 0.147m^2\times \sin0\\ &= \small 0 \end{aligned}$

• Always consider the field/resolution of it which is perpendicular to the plane.