Question

A uniform electric field of magnitude 7.0 ✕ 104 N/C passes through the plane of a square sheet with sides 5.0 m long. Calculate the flux (in N · m2/C) through the sheet if the plane of the sheet is at an angle of 60° to the field. Find the flux for both directions of the unit normal to the sheet.

Answer 1

Answer:

a) $\Phi=8.75*10^{5} [Nm^{2}/C]$

b) $\Phi=-8.75*10^{5} [Nm^{2}/C]$    

Explanation:

Let's recall the definition of the flux:

$\Phi=\textbf{E}\cdot \textbf{A}=|E||A|cos(\theta)$

$\Phi=\textbf{E}\cdot \textbf{A}=|E||L^{2}|cos(\theta)=7*10^{4}*5^{2}cos(60)$  

Therefore, the flux in the direction of the unit normal to the sheet is:

$\Phi=8.75*10^{5} [Nm^{2}/C]$

And the flux in the opposite direction is:

$\Phi=7*10^{4}*5^{2}cos(120)$

$\Phi=-8.75*10^{5} [Nm^{2}/C]$    

I hope it helps you!    

Answer 2

Answer:

$1.52*10^6 Nm^2/C$

Explanation:

Given that:

Electrical field E = $7.0 * 10^{-4}N/C$

square side l = 5.0 m

Area A = 5.0 * 5.0

= 25.0 m²

Angle ( θ ) between area vector and E = (90° - 60°)

= 30°

The flux $\phi_E$ can now be determined by using the expression

$\phi_E$ = $E*A*Cos \theta$

$\phi_E$ = $7.0 * 10^{-4}N/C$ $*25.0m*Cos 30^0$

$\phi_E$ = $1515544.457 Nm^2/C$

$\phi_E$ = $1.52*10^6 Nm^2/C$