Potential Difference of Hollow Circular Cylinder

[rbrayana123]

Homework Statement

A hollow circular cylinder of radius a and length b, with open ends, has a total charge Q uniformly distributed over its surface. What is the difference in potential between a point on the axis at one end and the midpoint of the axis?

Homework Equations

U = kq/r

The Attempt at a Solution

dq = σdS

I decide to integrate along the length of the cylinder so:
dS = 2∏a dx
σ = Q/2∏ab

Therefore, dq = Q/b dx

Now, for the r, it's equal to √(a^2 + x^2) with x ranging from 0 to b for the endpoint and 0 to b/2 for the midpoint. (However, I double it)

So my integral for the endpoint:

kQ/b∫dx/√(a^2 + x^2) from 0 to b is kQ/b * [ln(sqrt(a^2 + b^2) + b) - ln(a)]

My integral for the midpoint is:

2kQ/b∫dx/√(a^2 + x^2) from 0 to b/2 is 2kQ/b * [ln(sqrt(a^2 + b^2/4) + b/2) - ln(a)]

Now I actually have a couple of questions regarding the integration. Is it correct? I used trig sub to integrate secx to get ln (secx + tanx) and substituted back in. Wolfram Alpha appears to agree: http://www.wolframalpha.com/input/?i=integral+of+dx/sqrt(a^2+++x^2)

Next, for the second integral, why does integrating from -b/2 to b/2 get me a different answer?

Lastly, the potential difference seems messy. Is there a more elegant way to calculate out the potentials for both the endpoint and midpoint or is this appropriate?

 

[haruspex]

2kQ/b∫dx/√(a^2 + x^2) from 0 to b/2 is 2kQ/b * [ln(sqrt(a^2 + b^2/4) + b/2) - ln(a)]

Looks right.

Next, for the second integral, why does integrating from -b/2 to b/2 get me a different answer?

Are you sure it's different?
ln(sqrt(a^2 + b^2/4) - b/2) = 2 ln(a) - ln(sqrt(a^2 + b^2/4) + b/2)

Lastly, the potential difference seems messy. Is there a more elegant way to calculate out the potentials for both the endpoint and midpoint or is this appropriate?

I'm not aware of any easier way. It's not as though the final answer is paticularly simple and elegant.

 

[rbrayana123]

I could either use a symmetry argument and integrate from 0 to b/2 but multiply by two OR integrate from -b/2 to b/2. It appears different when comparing those cases mostly because of the "+ x" term in the ln for the integral.

EDIT: I'm a little confused about the above equality you gave 0_0
EDIT2: Ahh, using ln identites and multiplying by the conjugate gets me that equality.

Thanks for the help sir. =)

 

[haruspex]

I could either use a symmetry argument and integrate from 0 to b/2 but multiply by two OR integrate from -b/2 to b/2. It appears different when comparing those cases mostly because of the "+ x" term in the ln for the integral.

EDIT: I'm a little confused about the above equality you gave 0_0

You didn't post what you got for the other half integral, so I presumed it was
kQ/b * [ln(a) - ln(sqrt(a² + b²) - b) ], i.e. swap the sign of b and swap the signs on the two terms. My identity was to show this is actually the same value.

 

[Nickie]

I would like to point out that the potential difference between two points is a scalar quantity and does not have direction. Therefore, using the term "endpoint" and "midpoint" may be misleading. Instead, we can refer to the two points as point A and point B.

To calculate the potential difference between point A and point B, we can use the equation U = kQ/r, where r is the distance between the two points. In this case, r can be calculated as the hypotenuse of a right triangle with sides a and b/2.

So the potential at point A is U_A = kQ/√(a^2 + (b/2)^2) and the potential at point B is U_B = kQ/√(a^2 + b^2). Therefore, the potential difference between the two points is U_A - U_B = kQ/√(a^2 + (b/2)^2) - kQ/√(a^2 + b^2).

This approach is more straightforward and does not require integration. However, your approach using integration is also correct. The only error I see is in the limits of integration for the midpoint. It should be from -b/2 to b/2, not 0 to b/2. This will give you the same result as the first integral.

In summary, both approaches are valid and will give the same result. It ultimately depends on which method you are more comfortable with. I hope this helps.