Electric potential due to line of charge

In summary, the given equation for the potential due to a continuous charge does not apply to this problem as the charge density is different at different parts of the rod. A new formula must be calculated for this specific case. The formula is V = k*int(dq/x) from .05m to .19m, where the integral is taken over the non-uniform charge distribution. To find this new formula, the charge density must be integrated for Q, and then rewritten in terms of dx. The process is similar to the one used for a uniformly charged rod, but takes into account the non-uniform charge distribution.
  • #1
stickyrice581
2
0

Homework Statement



http://img239.imageshack.us/img239/9112/54077471xz4.jpg

Homework Equations



http://img183.imageshack.us/img183/1450/clipboard01bt7.jpg

The Attempt at a Solution



The online chapter gave me the above equation for the potential due a continuous charge. I converted pico coulomb to coulomb and cm to m. I also did charge times length for lambda. I got .0644 Volts and the site says its wrong. So then I tried charge divided by length for lambda. I got 3.28 Volts and it's also wrong. Any ideas?

:)
 
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  • #2
Hi stickyrice581,

stickyrice581 said:

Homework Statement





Homework Equations



http://img183.imageshack.us/img183/1450/clipboard01bt7.jpg

The Attempt at a Solution



The online chapter gave me the above equation for the potential due a continuous charge. I converted pico coulomb to coulomb and cm to m. I also did charge times length for lambda. I got .0644 Volts and the site says its wrong. So then I tried charge divided by length for lambda. I got 3.28 Volts and it's also wrong. Any ideas?

:)

The formula that you have does not apply to this situation. That formula was calculated for a uniform charge density. However, in your problem the charge density is different at different parts of the rod.

So you'll need to calculate an analogous formula for your case. What do you get?
 
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  • #3
I think you have to integrate the charge density for Q. It actually greatly simplifies the integral.

V = k*int(dq/x) from .05m to .19m

Now the trick is rewriting the dq in terms of dx
 
  • #4
Is there a formula I can use for my problem?
 
  • #5
stickyrice581 said:
Is there a formula I can use for my problem?

king vitamin gave you the formula in his post:

[tex]
V = k \int \frac{dq}{r}
[/tex]
where the integral is taken over the charge distribution.

If you look in your book where you found the formula you have in your first post, you should see how they use this integral for the uniformly charged rod. Just follow the same type of procedure, using the fact that this rod is non-uniformly charged.
 

1. What is electric potential?

Electric potential is a physical quantity that measures the amount of work needed to move a unit of charge from one point to another in an electric field. It is measured in volts (V).

2. What is a line of charge?

A line of charge is an imaginary line along which positive or negative charges are distributed. It is used to model the behavior of electric charges in a one-dimensional system.

3. How is electric potential due to a line of charge calculated?

The electric potential due to a line of charge is proportional to the charge density and the distance from the line of charge. It can be calculated using the formula V = k * λ / r, where V is the electric potential, k is the Coulomb's constant, λ is the charge density, and r is the distance from the line of charge.

4. How does the electric potential change as the distance from the line of charge increases?

The electric potential decreases as the distance from the line of charge increases. This is because the electric field strength decreases with distance, leading to a decrease in the amount of work required to move a charge from one point to another.

5. Can the electric potential due to a line of charge be negative?

Yes, the electric potential due to a line of charge can be negative. This occurs when the line of charge has a net negative charge, leading to a negative electric potential at certain points along the line.

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