Electric flux through a surface

In summary, the electric flux through the plane surface can be found by using the equation EAcosθ, where E is the electric field, A is the area of the surface, and θ is the angle between the electric field and the normal to the surface. In this case, with a uniform electric field of 375 N/C, an area of 0.0472 m2, and an angle of 64.2°, the calculated answer is 0.167 N·m2/C. However, it was discovered that the calculator was in radian mode, resulting in an incorrect answer. After correcting the mode and recalculating, the correct answer was found to be approximately 0.3605 N·m2
  • #1
haha1234
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0

Homework Statement



Find the electric flux through the plane surface shown in the figure below if θ = 64.2°, E = 375 N/C, and d = 4.70 cm. The electric field is uniform over the entire area of the surface.

Homework Equations





The Attempt at a Solution



By using the equation:EAcosθ
375x0.0472cos64.2=0.167
I've found that answer is 0.167N.m2/C,but it is not correct.
What is the correct answer?
 

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  • #2
Please show us exactly how you calculated your answer.
 
  • #3
jtbell said:
Please show us exactly how you calculated your answer.

I'v edited the content of my thread.Tell me of there are any things are need to be added.
 
  • #4
haha1234 said:
I'v edited the content of my thread.Tell me of there are any things are need to be added.

You have done the calculations incorrectly, what did you input in your calculator?
 
  • #5
Pranav-Arora said:
You have done the calculations incorrectly, what did you input in your calculator?

375x0.0472cos64.2=0.167
 
  • #6
haha1234 said:
375x0.0472cos64.2=0.167

I don't seem to be getting 0.167 with Wolfram Alpha.
 
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  • #7
Pranav-Arora said:
I don't seem to be getting 0.167 with Wolfram Alpha.

But I cannot get he other answer.:cry:
 

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  • #8
E=375
d2 = .002209
Cos64.2° = .435231

Multiply the three quantities and you should get something like .3605
 
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  • #9
haha1234 said:
But I cannot get he other answer.:cry:

because the tool is interpreting the angle 64.2 in radians .
 
  • #10
Tanya Sharma said:
E=375
d2 = .002209
Cos64.2° = .435231

Multiply the three quantities and you should get something like .3605

Thanks. I finally found that I have changed the mode of my calculator into radian,so I cannot find the correct answer.
 

FAQ: Electric flux through a surface

What is electric flux through a surface?

The electric flux through a surface is a measure of the amount of electric field passing through that surface. It is represented by the symbol ΦE and is measured in units of volts per meter (V/m).

How is electric flux through a surface calculated?

The electric flux through a surface is calculated by taking the dot product of the electric field and the surface area vector. This can be expressed mathematically as ΦE = E•A, where E is the electric field and A is the surface area vector.

What is the significance of electric flux through a surface?

The electric flux through a surface is a useful concept in understanding electric fields and their effects. It can help us determine the strength and direction of an electric field, as well as the potential difference between two points in the field.

How does the orientation of a surface affect electric flux?

The orientation of a surface can greatly affect the electric flux through it. If the surface is perpendicular to the electric field, the flux will be at its maximum value. However, if the surface is parallel to the electric field, the flux will be zero.

What are some real-world applications of electric flux through a surface?

Electric flux through a surface is used in many practical applications, such as calculating the capacitance of a capacitor, determining the strength of an electric field in a given region, and analyzing the behavior of charged particles in an electric field.

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