Electrical Potential of Semicircle Insulating Wire

In summary, the conversation discusses the calculation of potential at the center of a semicircular wire with a uniformly distributed charge. The speaker integrated the equation and arrived at the same answer as the book's solution manual. However, the book used a longer method involving breaking down dq and integrating from zero to pi. The speaker is unsure why this is necessary and seeks clarification. The responder explains that the speaker's method is valid for finding potential, but not for finding electric field at the center.
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Homework Statement



A thin insulating rod is bent into a semicircular wire of radius a, with a total charge of Q distributed uniformly along the rod. Calculate the potential at the center of the curvature if the potential is assumed to be zero at infinity.


Homework Equations



V=ʃ (kdq/r)



The Attempt at a Solution



I integrated the equation from zero to Q, and my final answer was kQ/a, which was the same as the book's solution manual. My problem is, the solution manual did it a completely different and longer way, and I'm just wondering if my method is not valid (because the book does it the same way as me on some similar examples).

the book made dq=(lambda)dl, where lambda=Q/pi*a, and dl=a*dθ.
Then, after moving stuff around, the final integral looked like this:

integral(k*Q/pi*a)dθ, and they integrated from zero to pi.


I've spent a few hours trying to figure out why you can just integrate dq from zero to Q in some instances, while in other instances that seem very similar (like this) you have to make dq equal something. Or, did the book do this as a matter of preference, and it is okay to integrate dq from zero to Q in this instance? Please help clear up this extreme fogginess for me. Thank you in advance to anyone who can help me.
 
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  • #2
In the case of potential you can use your method. Because the net potential is the scalar addition of potential due to dq. But if you want to find the electric field at the center, you cannot use this method.
 

FAQ: Electrical Potential of Semicircle Insulating Wire

What is the electrical potential of a semicircle insulating wire?

The electrical potential of a semicircle insulating wire depends on several factors such as the material of the wire, the length of the wire, and the amount of charge on the wire. It can be calculated using the formula V = kQ/r, where V is the potential, k is the Coulomb's constant, Q is the charge, and r is the radius of the semicircle.

How is the electrical potential of a semicircle insulating wire affected by the material of the wire?

The electrical potential of a semicircle insulating wire is affected by the material of the wire because different materials have different levels of conductivity. Conductive materials allow for easier flow of electricity, resulting in a higher potential, while insulating materials have higher resistance and therefore lower potential.

Does the length of the semicircle insulating wire affect its electrical potential?

Yes, the length of the semicircle insulating wire does affect its electrical potential. The longer the wire, the higher the potential, as there is more surface area for the charge to accumulate. This is why longer wires are used for transmitting electricity over longer distances.

Can the electrical potential of a semicircle insulating wire be negative?

Yes, the electrical potential of a semicircle insulating wire can be negative. This can occur when there is an excess of negative charge on the wire, resulting in a negative potential. However, in most cases, the potential of a semicircle insulating wire is positive.

How does the amount of charge on a semicircle insulating wire affect its electrical potential?

The amount of charge on a semicircle insulating wire directly affects its electrical potential. The higher the charge, the higher the potential. This is because the potential is directly proportional to the charge and inversely proportional to the radius of the semicircle.

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