4.3. Line of Charge and Two Conducting Walls In the previous problem set, you we

Question

4.3. Line of Charge and Two Conducting Walls In the previous problem set, you were tolo "The well-known potential for an isolated line charge at (z ) is ?(z, y) Derive this expression using 1. Gauss' law and (A/4?? ) n (R2/r2), where r2 (z z.? + (y-y )2 and R is a constant." For the second case, use the delta function notation for ? to show how ? d3z reduces to ?dz. Make sure that your notation is technically correct. Also consider what the notation would be for a finite line of charge. Sketch the equipotential lines. You should be able to reason this out before even calculating the potential. If you did not get the algebra to work for ? ? Pasym, calculate it along the line z = y again assuming zo-Yo. (For a problem with many algebraic steps, you should probably do this anyway as there are fewer opportunities for error and the result can be used to check the general result.) When you calculated the charge density for the case zo -yo, what was the location of the minimum charge density? Does this match what you would expect physically? Can you think of other features the solution for ?/X should have based on physical reasoning that you can use to check your ca a on?

Explanation / Answer

1) Given, isolated line charge in xy – plane passing through (xo,yo)

Charge per unit length = ?

We know that, if we consider a cylinder which is centred at this line charge. The electric field is constant throughout the curved surface and is in the direction of area vector (equipotential surfaces). Also, the electric field at the two flat surfaces is zero. Because of this symmetry we can use gauss law to find electric feild.

Let us find the electric field at an arbitrary point (x,y). To make this simpler let us assume that this point is on the perpendicular to line charge at (xo,yo).

Radius of cylinder centred at (xo,yo) and passing through (x,y) is r = ((x-xo)2 + (y-yo)2)1/2

Length of cylinder be ‘l’

Let ‘R’ be the radius of cylinder having reference potential i.e. zero. Generally this is assumed to be at infinity.

We know that, from gauss law

                ? E.da = Q/?o (Q = total charge enclosed in the cylinder = ?l)

As the electric field is constant all over the surface of integration and is along the direction of da . Therefore, E can be taken out of integral.

                E ?(over the curved surface) da = ?l/ ?o

The value of integral is just the area of curved surface ‘A’ = 2?rl

                2?rlE = ?l/ ?o

                E = ?/2??or s^

Direction of E is radially outward i.e. s^

We know that, E = - del (V) where V is potential.

As we need only the s component of Del operator which is d/ds. Let us replace s by r to avoid confusion.

                dV/dr = -E

                dV = - ?/2??or dr

                By integrating on both sides from R to r we get

                V = - ?/2??o ln(r/R)

                By absorbing – sign onside r and multiplying and dividing by 2 we get the required form

                V = ?/4??o ln(R2/r2)

2) Given, d3x as volume element which is equivalent to dxdydz and ? is the charge density in three dimensions.

But, we have line charge which is a one dimensional element whose density is ?.

So, by using delta functions we can write ? in terms of ? as

                ? = ? ??2(x)

Where, ??2(x) is equivalent to ??(y) ??(z) and the line charge is along x- axis.

By, substituting this ? in the given equation d3x reduces to dx. (? ??(x)dx =1 )

3) equipotential surfaces are just cylinders which are centered at the line charge.

4) ?asym definition has not been specified.

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