Electric potential, Resistance , Capacitor

In summary: GIn summary, the conversation discusses a problem involving a charge line and a conductive cylinder attached to the Earth, with a request for help in determining the capacity of the system as a capacitor. The potential and field equations are provided, along with a suggestion to consider the rearrangement of charges in the conductor for solving the problem.
  • #1
a_aziy
2
0
Hi friends
Thanks for your helps... :smile:
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 ... :wink:
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 ?
See attachment for more information ...
 

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  • #2
a_aziy said:
Hi friends
Thanks for your helps... :smile:
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 ... :wink:
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 ?
See attachment for more information ...
here is a start:

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

[tex]V = \int_d^\infty E\cdot dr[/tex]

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

[tex]\int E\cdot dA = E 2\pi r L = q/\epsilon_0 = \lambda L/\epsilon_0[/tex]

So:
[tex]E = \frac{\lambda}{2r\pi\epsilon_0}[/tex]

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


Use [itex]C=\frac{Q}{V}[/itex].

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
 
  • #3


Sure, I'd be happy to help! So, to start off, let's define electric potential, resistance, and capacitor. Electric potential is the amount of electric potential energy per unit of charge at a specific point in an electric field. It is measured in volts (V). Resistance is the measure of how difficult it is for current to flow through a material. It is measured in ohms (Ω). A capacitor is a device that stores electric charge and is made up of two conductors separated by an insulator. It is measured in farads (F).

Now, let's address the question about the capacitor with a charge line and a conducting cylinder attached to the Earth. To find the capacitance of this system, we can use the formula C = Q/V, where C is the capacitance, Q is the charge, and V is the voltage. In this case, the voltage is the difference in electric potential between the two conductors.

Since the charge line has an infinite length, we can assume that the charge is evenly distributed along its length. This means that the charge density (λ) is equal to the total charge (Q) divided by the length (l) of the line. In other words, λ = Q/l.

Now, to find the voltage between the two conductors, we can use the formula V = Ed, where E is the electric field strength and d is the distance between the two conductors. We can find E by using the formula E = λ/2πεd, where ε is the permittivity of the material between the two conductors.

Putting all of this together, we get C = λd/2πε, which is the capacitance of this system. I hope this helps! Let me know if you have any other questions or if you need further clarification.
 

1. What is electric potential and how is it measured?

Electric potential is the amount of electrical energy that a charged particle has due to its position in an electric field. It is measured in units of volts (V) using a voltmeter.

2. What is resistance and how does it impact the flow of electric current?

Resistance is the measure of how difficult it is for electric current to flow through a material. It is measured in units of ohms (Ω). A higher resistance means there is less current flow, while a lower resistance allows more current to flow.

3. How does a capacitor store and release electric charge?

A capacitor is an electronic component that stores electric charge. It consists of two conductive plates separated by an insulating material. When a voltage is applied, one plate becomes positively charged and the other becomes negatively charged. This stored charge can then be released when the capacitor is connected to a circuit.

4. How can the capacitance of a capacitor be increased?

The capacitance of a capacitor can be increased by changing the area of the plates, increasing the distance between them, or using a material with a higher dielectric constant. Additionally, connecting capacitors in series or parallel can also increase the overall capacitance.

5. What factors affect the electric potential difference between two points?

The electric potential difference, or voltage, between two points is affected by the amount of charge present, the distance between the points, and the type of material in between. Additionally, the presence of other electric fields or sources can also impact the potential difference.

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