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denpristay [2]
3 years ago
14

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.
Physics
2 answers:
Anika [276]3 years ago
7 0

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!    

Vadim26 [7]3 years ago
5 0

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

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