Conservative Nature of Electric forces

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Homework Help Overview

The discussion revolves around the conservative nature of electric forces, particularly in the context of static charge distributions and their effects on charged particles. Participants are exploring the implications of energy conservation when a charged particle is accelerated by an electric field.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants are questioning the relationship between conservative forces and energy dissipation, particularly in the context of electromagnetic radiation when a charge is accelerated. There is a focus on whether energy conservation equations can be applied in this scenario.

Discussion Status

Some participants have provided definitions of conservative forces and discussed the implications of energy loss during acceleration. There is an ongoing exploration of the definitions and properties of conservative forces, with no explicit consensus reached on the assumptions being made regarding energy dissipation.

Contextual Notes

Participants are considering the assumption that energy loss due to radiation is negligible in the context of the problem, which may affect the application of energy conservation principles.

SpartanG345
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My textbook says for every electric field due to a static charge distribution, the force exerted by that field is conservative.

If that force is used to accelerate a charged particle, wouldn't the particle disapate some of its energy, as EMR? therefore you cannot say U_a + K_a = U_b + K_b as some of the particles energy would have gone

can anyone explain what that above statement means with respect to the motion of the particle?

thankyou :)
 
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SpartanG345 said:
My textbook says for every electric field due to a static charge distribution, the force exerted by that field is conservative.

If that force is used to accelerate a charged particle, wouldn't the particle disapate some of its energy, as EMR? therefore you cannot say U_a + K_a = U_b + K_b as some of the particles energy would have gone

can anyone explain what that above statement means with respect to the motion of the particle?

thankyou :)

I've highlighted the key word in your textbook's statement...In electrostatics charges don't radiate.
 
i understand that, the charge distribution is static, but there was a question, where a charge is accelerated by a static charge distribution.

Say a charge is being accelerated by a static charge distribution, wouldn't that charge lose some energy?

such that is incorrect to say Ua + Ka = Ub + Kb, since the charge accelerates
 
Sure, the accelerating charge will radiate some energy away. However, this is unimportant. Energy conservation and conservative forces are not the same thing. What is the actual definition of a conservative force?
 
A conservative force is a force that conserves energy, such that the work done by a conservative force can be released and is reversible.

So i suppose as the electron moves away some of its kinetic energy is dissipated as EMR but its Kinetic energy is also turned into potential energy which can be recovered.

But it is still wrong to say, Ua + Ka = Ub + kb as some energy is lost in the movement from a to b, i think the question assumes that this loss is negligible don't you think?
 
SpartanG345 said:
A conservative force is a force that conserves energy
That's a somewhat inaccurate definition.

the work done by a conservative force can be released and is reversible.

This is a much better definition...From this definition, you can say 3 mathematically equivalent things:

(1)The work done by the force over any simple, closed path is zero (otherwise it would be an irreversible process)

W=\oint\textbf{F}\cdot d\textbf{r}=0

(2)The curl of \textbf{F} must vanish (you can derive this easily using Stokes's Theorem and property 1)

\mathbf{\nabla}\times{\textbf{F}}=0

(3)The force can be written as the (negative by convention) gradient of a single-valued, contiuous potential function

\textbf{F}=-\mathbf{\nabla}UNow, surely any electrostatic force satisfies property 2 (and hence the other two properties as well), right?
 

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