Answer:
[tex]1.52*10^6 Nm^2/C[/tex]
Explanation:
Given that:
Electrical field E = [tex]7.0 * 10^{-4}N/C[/tex]
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 [tex]\phi_E[/tex] can now be determined by using the expression
[tex]\phi_E[/tex] = [tex]E*A*Cos \theta[/tex]
[tex]\phi_E[/tex] = [tex]7.0 * 10^{-4}N/C[/tex] [tex]*25.0m*Cos 30^0[/tex]
[tex]\phi_E[/tex] = [tex]1515544.457 Nm^2/C[/tex]
[tex]\phi_E[/tex] = [tex]1.52*10^6 Nm^2/C[/tex]
Answer:
a) [tex]\Phi=8.75*10^{5} [Nm^{2}/C][/tex]
b) [tex]\Phi=-8.75*10^{5} [Nm^{2}/C][/tex]
Explanation:
Let's recall the definition of the flux:
[tex]\Phi=\textbf{E}\cdot \textbf{A}=|E||A|cos(\theta)[/tex]
[tex]\Phi=\textbf{E}\cdot \textbf{A}=|E||L^{2}|cos(\theta)=7*10^{4}*5^{2}cos(60)[/tex]
Therefore, the flux in the direction of the unit normal to the sheet is:
[tex]\Phi=8.75*10^{5} [Nm^{2}/C][/tex]
And the flux in the opposite direction is:
[tex]\Phi=7*10^{4}*5^{2}cos(120)[/tex]
[tex]\Phi=-8.75*10^{5} [Nm^{2}/C][/tex]
I hope it helps you!
