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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.

Respuesta :

ajeigbeibraheem

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]

whitneytr12

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!    

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