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auk411
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Homework Statement
see below.
Homework Equations
The Attempt at a Solution
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Electric Field: Homework Solutions
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In summary, the conversation discusses the calculation of the electric field above two non-conducting sheets with identical distributions of positive charge. The answer is \sigma/epsilon\epsilon0 due to the principle of superposition. However, the initial attempt at the solution resulted in the incorrect answer of zero, which was found to be wrong. The correct answer is \sigma/epsilon\epsilon0 as the negative charges in the image are not supposed to be there.
see below.
- #2
auk411
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See the attachment for the visual, but ignore the negative charges. They aren't supposed to be there.
The image is of two large non-conducting sheets with identical distributions of positive charge. What is the value of the electric field above the sheets? sigma is the symbol for the identical distribution of positive charge.
I know the answer: [itex]\sigma[/itex]/epsilon[itex]\epsilon0[/itex]
I know that you need to add to use the principle of superposition.
However I still don't get the answer. The top sheet adds [itex]\sigma[/itex]/2epsilon[itex]\epsilon0[/itex] upwards. The downwards from the top sheet is canceled out by the upwards (electric field) from the bottom sheet. The bottom sheet has an electric field that also points in the negative y direction but with the same magnitude. So everything cancels out. So the electric field is zero.
This is wrong. However, I still do not see why and why the right answer is right.
The image is of two large non-conducting sheets with identical distributions of positive charge. What is the value of the electric field above the sheets? sigma is the symbol for the identical distribution of positive charge.
I know the answer: [itex]\sigma[/itex]/epsilon[itex]\epsilon0[/itex]
I know that you need to add to use the principle of superposition.
However I still don't get the answer. The top sheet adds [itex]\sigma[/itex]/2epsilon[itex]\epsilon0[/itex] upwards. The downwards from the top sheet is canceled out by the upwards (electric field) from the bottom sheet. The bottom sheet has an electric field that also points in the negative y direction but with the same magnitude. So everything cancels out. So the electric field is zero.
This is wrong. However, I still do not see why and why the right answer is right.
Attachments
1. What is an electric field?
An electric field is a physical quantity that describes the influence exerted by electric charges on other charges in their vicinity. It is represented by a vector quantity, with both magnitude and direction, and is typically measured in units of volts per meter (V/m).
2. How is the strength of an electric field determined?
The strength of an electric field is determined by the magnitude of the electric charges and their distance from each other. The closer the charges are, the stronger the electric field will be. The strength of the electric field also depends on the medium in which the charges are located, such as air, water, or vacuum.
3. What is the difference between an electric field and an electric potential?
An electric field describes the force experienced by a charged particle in the presence of other charges, while electric potential describes the energy required to move a charged particle from one point to another in an electric field. In other words, the electric field is the cause and the electric potential is the effect.
4. How can electric field lines be used to visualize the electric field?
Electric field lines are imaginary lines that represent the direction and strength of the electric field. They are drawn such that the tangent to the line at any point gives the direction of the electric field at that point. The closer the lines are to each other, the stronger the electric field at that point.
5. How can I use Coulomb's Law to calculate the electric field?
Coulomb's Law states that the electric field between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. It can be used to calculate the electric field at a point due to one or more charges by summing up the individual contributions from each charge using vector addition.
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