- #1

vcsharp2003

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- Homework Statement
- Imagine a cylinder that is symmetrical about an infinite plane sheet of charge as shown in the diagram. What is the electric flux through the ends ##A \text{ and } A^{'}## of this imaginary cylinder if ##\Delta {S}## is the area of each of these ends?

- Relevant Equations
- ##\phi_E = \oint \vec E \cdot \vec {dS}##, which is the definition of electric flux

When I look at this question, I can see two possible values of electric flux depending on how I take the normal area vector for either ends ##A \text{ and } A^{'}##.

What is wrong with my logic below where I am ending up with two possible answers? The book mentions that only ##2E\Delta{S}## is correct.

We know that electric field will be the same at each and every point on ##A \text{ and } A^{'}##due to the symmetry of charge distribution on the plane sheet and it will always be perpendicular to ##A \text{ and } A^{'}##.

Now consider an infinitesimal area ##dS## on each of these surfaces.

Then electric flux through ##A## could be one of the two values since the normal area vector could be either towards the infinite plane sheet or away from the sheet: ##\phi_E = \oint \vec E \cdot \vec {dS} = \oint E dS \cos{0} \text{ , } \oint E {dS} \cos{2\pi} ##

So, electric flux through ##A## could be either ##E\Delta{S}## OR ##-E\Delta{S}##. The minus sign in second value is because ##\cos{2\pi} = -1##

From above it follows that electric flux through the ends ##A \text{ and } A^{'}## could be ##2E\Delta{S}## or ##0##. But the answer given is the first value and not the second value.

What is wrong with my logic below where I am ending up with two possible answers? The book mentions that only ##2E\Delta{S}## is correct.

We know that electric field will be the same at each and every point on ##A \text{ and } A^{'}##due to the symmetry of charge distribution on the plane sheet and it will always be perpendicular to ##A \text{ and } A^{'}##.

*So, let's assume that this electric field at every point of these surfaces is E.*Now consider an infinitesimal area ##dS## on each of these surfaces.

Then electric flux through ##A## could be one of the two values since the normal area vector could be either towards the infinite plane sheet or away from the sheet: ##\phi_E = \oint \vec E \cdot \vec {dS} = \oint E dS \cos{0} \text{ , } \oint E {dS} \cos{2\pi} ##

So, electric flux through ##A## could be either ##E\Delta{S}## OR ##-E\Delta{S}##. The minus sign in second value is because ##\cos{2\pi} = -1##

From above it follows that electric flux through the ends ##A \text{ and } A^{'}## could be ##2E\Delta{S}## or ##0##. But the answer given is the first value and not the second value.

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