Question about equipotential lines and the work done moving along them

AI Thread Summary
Equipotential surfaces have the same electric potential, meaning no work is needed to move a charge between points on these surfaces. The discussion clarifies that while no net work is required, a small amount of force can be applied to change a particle's location without accelerating it. The position vector "r" is defined in Cartesian coordinates, and the force acting on a particle can be expressed as the gradient of the potential. It is noted that while the electric field does no work along equipotential lines, other forces can still act on the particle. Understanding these concepts is crucial for grasping the dynamics of particles in electric fields.
engineeringstudnt
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Homework Statement
electric potential
Relevant Equations
v=Kq/r
hi guys i have a conceptual question .As you know equipotential surfaces is one on which all point are the same potential there is no work required to move a charge from one point to the other . So my question is how can we change the locotion of a particle without using any force ?
 
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The correct formulation is as follows. If a particle moves such that ##V(\boldsymbol r(t))=const## then the force $$\boldsymbol F=-\frac{\partial V}{\partial \boldsymbol r}$$ does no work.
 
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engineeringstudnt said:
Homework Statement:: electric potential
Relevant Equations:: v=Kq/r

hi guys i have a conceptual question .As you know equipotential surfaces is one on which all point are the same potential there is no work required to move a charge from one point to the other . So my question is how can we change the locotion of a particle without using any force ?
You are assuming a particle with mass, yes? And that there is no gravitational field, or the gravitational equipotential coincides with the electrostatic one?
No net work is required. Perhaps the particle is orbiting at constant speed, or you may put some arbitrarily small amount of work into accelerate it then extract that work to stop it.
 
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haruspex said:
You are assuming a particle with mass, yes? And that there is no gravitational field, or the gravitational equipotential coincides with the electrostatic one?
No net work is required. Perhaps the particle is orbiting at constant speed, or you may put some arbitrarily small amount of work into accelerate it then extract that work to stop it.
i guess i understand. so we should still use some force even its very very small.if particle is not accelarated initially. right ?
 
wrobel said:
The correct formulation is as follows. If a particle moves such that ##V(\boldsymbol r(t))=const## then the force $$\boldsymbol F=-\frac{\partial V}{\partial \boldsymbol r}$$ does no work.
thanks sir but can i ask what is "r"?
 
engineeringstudnt said:
thanks sir but can i ask what is "r"?
I think its the position vector of the particle, in cartesian coordinates ##r=x\hat x+y\hat y+z\hat z## where x the x-coordinate of the particle and ##\hat x## the unit vector of the cartesian system in the x-axis. Also I think the symbolism used it just translates to $$F=-\nabla V=-\frac{\partial V}{\partial x}\hat x-\frac{\partial V}{\partial y}\hat y-\frac{\partial V}{\partial z}\hat z$$.
 
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Thank you all :)
 
As for your main question in equipotential lines the work of the field under consideration is zero, we can do work on the particle with other forces from other fields that don't share the same equipotential curves.
 
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A force normal to the particle's velocity does no work yet would cause a deflection.
 
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