Electric potential, Resistance , Capacitor

a_aziy

Hi friends
Thanks for your helps... but still I have problem with these 3 question ...
I've attached a picture which contains questions and drawing of each one...
I will be so pleased ,if you help me ... 1) There is a charge line with a length of infinite and charge density of (landa) and distance of (d) from center of a conductive cylinder which is attached to the earth , What is the capacity of this system as a capacitor ?

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Andrew Mason

Homework Helper
a_aziy said:
Hi friends
Thanks for your helps... but still I have problem with these 3 question ...
I've attached a picture which contains questions and drawing of each one...
I will be so pleased ,if you help me ... 1) There is a charge line with a length of infinite and charge density of (landa) and distance of (d) from center of a conductive cylinder which is attached to the earth , What is the capacity of this system as a capacitor ?
here is a start:

The potential at a distance d from the centre of the line charge is:

$$V = \int_d^\infty E\cdot dr$$

The field E is determined by Gauss' law taking a gaussian cylinder of length L around the line charge:

$$\int E\cdot dA = E 2\pi r L = q/\epsilon_0 = \lambda L/\epsilon_0$$

So:
$$E = \frac{\lambda}{2r\pi\epsilon_0}$$

$$V = \int_d^\infty \frac{\lambda}{2r\pi\epsilon_0}\cdot dr = -\frac{ln(d)\lambda}{2\pi\epsilon_0}$$

Use $C=\frac{Q}{V}$.

Now the tricky part:

To work out the capacitance of the conducting cylinder, centered d distance from the line charge, one has to recognize that the charges in a conductor rearrange themselves in the presence of the electric field from the line charge so as to make all the charges in the conductor at equal potential. The fact that the conducting cyclinder is connected to ground means that the free charges in the cylinder, when subjected to the field from the line charge, are all at 0 potential.

Can determine what the charge must be on the conducting cyclinder in order to keep the cylinder at 0 potential in the presence of the line charge? If so, I think you have solved the problem.

AM

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