# Question about equipotential lines and the work done moving along them

engineeringstudnt
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 ?

• Delta2

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.

• engineeringstudnt
Homework Helper
Gold Member
2022 Award
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.

• Steve4Physics and engineeringstudnt
engineeringstudnt
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 ?

engineeringstudnt
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"?

Homework Helper
Gold Member
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 $$F=-\nabla V=-\frac{\partial V}{\partial x}\hat x-\frac{\partial V}{\partial y}\hat y-\frac{\partial V}{\partial z}\hat z$$.

• wrobel and engineeringstudnt
engineeringstudnt
Thank you all :)

Homework Helper
Gold Member
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.

Last edited:
e^(i Pi)+1=0
A force normal to the particle's velocity does no work yet would cause a deflection.

• Delta2