Coaxial Cylinders. A long metal cylinder with radius a is...

Coaxial Cylinders. A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius $b$. The positive charge per unit length on the inner cylinder is $\lambda$, and there is an equal negative charge per unit length on the outer cylinder. (a) Calculate the potential $V(r)$ for (i) $r < a$; (ii) $a < r < b$; (iii) $r > b$. ($Hint:$ The net potential is the sum of the potentials due to the individual conductors.) Take $V = 0$ at $r = b$. (b) Show that the potential of the inner cylinder with respect to the outer is $$V^{ab} = \frac{\lambda} {2\pi\epsilon_0} ln \frac{b} {a}$$ (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the cylinders has magnitude $$E(r) = \frac{V_{ab}} {ln(b/a)} \frac{1} {r}$$ (d) What is the potential difference between the two cylinders if the outer cylinder has no net charge?

Transcript

00:01 The problem is given to us the coaxial cylinders.

00:06 A long metal cylinder with a radius a is supported on an insulating stand on the axis of a long hollow metal tube with radius b.

00:21 The positive charge per unit length on the inner cylinder is lambda.

00:28 And there is an equal negative.

00:32 Charge per unit length on the outer cylinder for part a it is asked that calculate the potential b tau for sorry we are for first you can say r is less than a b for b when a is less than r is less than b and for third its r is greater than b so here hint is given to us that is the net potential is the sum of potential due to the individual conductors.

01:14 Take v equals to 0 at r equals to b and show that the potential of the inner cylinder with respect to the outer is vab equals to lambda by 2 epsilon 0 lnb by a.

01:31 And for part c it is asked, use equation 23 .23 .23 and the result from the part a.

01:38 To show that the electric field at any point between the cylinders has magnitude er equals to b a b by ln b by a into 1 by r now for d part it is said that what is the potential difference between the two cylinders if the outer cylinder has no net charge so now we will see here how to do so so the equation for potential outside the cylinder is b equals to lambda by 2 by absalom and then r not by r so here r not is the distance from the axis for which we take b equals to zero right so now r is the distance from the cylinder axis all right so for holo tube of radii b and charge per unit length equals to you can say this is this is lambda right length is lambda right so the potential outside the cylinder is v equals to lambda by 2 pi epsilon not l n b by r right so the potential inside the cylinder will be v equals to zero so v equals to zero at r and equals to b when where radius is b so b is the radii of what follow tube for the metal cylinder of radius a and charge per unit length lambda b is given by what lambda equal lambda upon 2 pi epsilon not l n b by r here now for inside v equals to lambda by 2 pi epsilon not l n b by a so when we are asked to find the potential v r you can see so this is at at r is less than a b is given by you can say lambda by 2 pi epsilon or not ln b by a minus of ln b by a here ln b by a and yes so it should be b by b i guess right right i should correct it it should be b by b here right so this becomes bb cutoffs and l n log 1 becomes 0 here so this finally makes lambda by 2 pi epsilon node, l and b by superior.

05:34 Now for a less than r, less than b, outside the cylinder, inside tube, outside the cylinder, but inside the tube, you can say, it becomes again lambda by 2 pi epsilon knot, lnb by r minus lnb by b.

05:57 So this comes out to be lambda by 2 pi epsilon knot lnb by ra...

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