What is the electric field strenght in units of N/C if the flux through a 2.0m by 1.0m rectangular surface is 396Nm2/C, if the electric field is uniform, and if the plane of the surface is at an angle of π/3 radians with respect to the direction of the field ?

Answers

Answer 1

Answer:

The appropriate answer is "396 N/C".

Explanation:

The give values are:

Rectangular surface,

[tex]\Phi_e=396 \ N.m^2/C[/tex]

[tex]a = 2 \ m[/tex]

[tex]b = 1 \ m[/tex]

Angle,

[tex]\theta =\frac{\pi}{3} \ radian[/tex]

Now,

The area of rectangle (A) will be:

=  [tex]a\times b[/tex]

=  [tex]2\times 1[/tex]

=  [tex]2 \ m^2[/tex]

hence,

The electric field strength will be:

⇒  [tex]\Phi_e=E.A Cos \theta[/tex]

or,

⇒  [tex]E=\frac{\Phi_e}{ACos \theta}[/tex]

On putting the given values, we get

⇒      [tex]=\frac{396}{2\times Cos(\frac{\pi}{3} )}[/tex]

⇒      [tex]=\frac{396}{2\times 0.5}[/tex]

⇒      [tex]=\frac{396}{1}[/tex]

⇒      [tex]=396 \ N/C[/tex]


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Answer:

Part A

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Answer:

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