Why electric potential of arc treated like point charge

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The discussion centers on understanding why the electric potential of an arc of uniform charge can be treated like that of a point charge, with the expression V = kQ/R derived from integrating contributions from individual charge elements. Participants clarify that electric potential is a scalar quantity, allowing contributions from different parts to be summed directly, unlike electric fields, which are vector quantities. The analogy to a conducting sphere is mentioned, emphasizing uniform charge distribution, but it is noted that the arc does not directly relate to a conducting sphere. The conversation also touches on the distinctions between electric force, electric field, electric potential, and electric potential energy, providing foundational context for these concepts. Overall, the treatment of the arc as a collection of point charges simplifies the calculation of electric potential at the center.
lonewolf219
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Homework Statement



Give an expression to find V of Arc of uniform charge (at the center, or origin)

Homework Equations



V=kQ/R


The Attempt at a Solution



the solution is kQ/R. I'm wondering why an arc can be treated like a point charge...
Is this reason partly connected to a conducting sphere, and the fact is has an equal charge distribution everywhere, even at its hollow center? Likewise, would a uniform arc also be seen as a hollow conducting sphere?
 
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hi lonewolf219! :smile:

electric potential is a scalar, and so we add the contributions (from different parts) as scalars (ie ordinary numbers) …

this works for any shape :wink:

(unlike electric field, which is a vector, and so we add the contributions as vectors)
 
Thanks Tiny-Tim...

Just starting the second semester of introductory physics. A little confusing with electric force, electric field, electric potential and electric potential energy...

As you pointed out, scalars versus vectors. But I guess there is little connection between an arc and a conducting sphere...
 
lonewolf219 said:

Homework Statement



Give an expression to find V of Arc of uniform charge (at the center, or origin)

the solution is kQ/R. I'm wondering why an arc can be treated like a point charge...

You can consider the arc as a lot of equal point charges arranged in arc from. A line element dL has a charge dQ, and contributes to the potential by dV= kdQ/R at the centre. The contributions of the charges add up: so the total potential at the centre is the integral of these contributions.

V=\int{\frac{kdQ}{R}}.

k/R is constant, it can be factored out from the integral, so V=\frac{k}{R}\int{dQ}=\frac{kQ}{R}.

ehild
 

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hi lonewolf219! :smile:

(just got up :zzz: …)
lonewolf219 said:
A little confusing with electric force, electric field, electric potential and electric potential energy...

electric force is a force (as in "F = ma")

electric field is electric force per charge

electric potential energy is what it says, the PE of the electric field

electric potential is electric potential energy per charge

(just as gravitational potential is gravitational potential energy per mass: PE/m = mgh/m = gh, or = -mMG/mr = -MG/r)
 
Thanks guys!

People like you are the reason physics forums is so awesome...
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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