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em-agbx-01001
| If the number of electrons per unit volume is \(n\), \(t\) is the thickness of the slab then the current through the slab is \( I=e v ne (w t)\), where \(w\) is the width and \(t\) is the thickness of the slab. WHY? |
If the current density is \(j\) the current flowing in the conductor is charge flowing per second along the \(X\)- axis.
The amount of charge flowing per sec across a slab, width =\(w\) will be that contained in volume of dimension \(w\times 1\times t\) where \(t\) is the thickness of the slab. Thus \(I=ne wt\).
em-agbx-02001
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Question : How to verify that the electric field due to point charge at rest obeys the Maxwell's second equation \(\nabla \times \vec{E}=0.\) |
By direct computation of curl of the electric field.
em-agbx-03001
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Question : How is path independence helpful.? |
The path independence property is helpful in many ways, for example the the line integral between two points can be computed along any path.
em-agbx-03002
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Question: When is true that the electric potential vanishes as \(r\to \infty\)? Give an example. |
Details:
If the charge distribution is confined to a finite volume, the potential will go to zero as \(r\to\infty\). This is a sufficient condition, but not necessary.
Basically the charge density must go to zero sufficiently fast when \(r\to\infty\). How fast? faster than \(1/r^\epsilon\), for some positive \(\epsilon\). This means for some \(R\)
\[ \rho(r) < \frac{\text{const}}{r^\epsilon}, \text{ if } r>R.\]
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4727:Diamond Point
