When moving a charge across an equipotential surface, why is the net force zero?

Question

In an equipotential surface, the work done is 0, because $W = q \cdot \Delta V$ and $\Delta V = 0$. But $W = Fs$ as well. As the displacement is non-zero, the net force in moving a charge, or a mass, across an equipotential surface is zero. I have two queries:

1. How exactly is the net force zero?
   
2. Does that suggest that charges/masses can only be moved across an equipotential surface at constant velocity? Wouldn't they have to be accelerated, even for a short amount of time, or does the attractive/repulsive force generated by the field somehow counter that (If so, how)?

Answer 1

It is important to realize that these are vectorial equations. The work is the integral of the force towards the path, along the path. More explicitly: your equation $W=Fs$ involves the scalar quantities "length of path" $s$ and "strength of force" $F$. It works well for a straight path, and a force in the same direction. But more generally, the path may be crooked and the force in any direction. So in each small segment of the path $\vec{ds}$ (vector!), the work along that segment is $\vec{F}\cdot\vec{ds}=F_{par}ds$ with $F_{par}$ being the component of the force in the direction of the path $\vec{ds}$. The work over the entire path is just the sum (integral) along the whole path, $$W=\int \vec{F}\cdot\vec{ds}$$.

The electric force created by the potential always points donwlope, and hence is always perpendicular to the equipotential surface. It thus has no component along the equipotential line, and thus contributes exactly zero to the integral.

Answer 2

An equipotential surface $\bf S$ is defined as the set of all points $\vec x\in \bf S$ for which the potential is equal to a specified constant value $$V(\vec x)=V_0=const.$$ If you have a parameterized curve $\vec x(\lambda)\in \bf S$ for every possible $\lambda$, i.e. the curve is "moving across the equipotential surface", then $$0 = \frac{d}{d\lambda}V_0=\frac{d}{d\lambda}V(\vec x(\lambda))=\frac{\partial V}{\partial \vec x}\frac{d\vec x}{d \lambda}$$ This is simply the chain rule of differentiation. Since the derivative of the potential is basically the force, i.e. $$\vec F=-\frac{\partial V}{\partial \vec x}$$ and the tangent vector to the curve at parameter $\lambda$ is given by $$\vec t(\lambda)=\frac{d\vec x}{d \lambda}/\left|\frac{d\vec x}{d \lambda}\right|$$ the derived equation reads $$0=\vec F(\lambda) \cdot \vec t(\lambda)=F_{tangent}$$ So the force component along the arbitrary curve (which has been constrained to the equipotential surface) is indeed zero.

You can of course also longitudinally accelerate a body constrained to an equipotential surface. But if the force that realizes the constraint is conservative, you might be inclined to include that force also into an encompassing potential. In this case, the former equipotential surface would not be "equipotential" for the extended potential anymore.

Answer 3

First of all, the equation $$W=q\Delta V$$ refers to the work done against/from the electric field that generated the potential. If you move on an equipotential surface, you simply do not push "against" the electric field nor does the field "push" against you.

So to answer the first question, it is not that the force is $0$ but rather that no contribution of the force due to the electric field is countering your displacing of the particle.

It is like moving a mass on a frictionless, flat surface (i.e. on an equipotential gravitational surface): weight does not matter. Gravity is there, but it is perpendicular to your displacement so for the total work you get $$W=\int \vec{F}\cdot \vec{ds}=0$$ (where $\vec{ds}$ is your displacement) because $\vec{F}\cdot \vec{ds}=0$ (on an equipotential surface, the force of the field is always parallel to the surface as the force is the gradient of the potential and the gradient on a constant surface is ... 0 so the force is also 0 on the plane - but it does not mean the total force is 0!).

If you set your mass in motion on an equipotential surface, it will keep moving (at constant speed, unless you provide extra force to it) as nothing is stopping/accelerating it. On the other hand, if you push the mass towards the top then you need to apply a counter-force continously to counter-act the force of gravity.

Notice that the work-energy theorem states that, in general, for a conservative force like gravity or electric forces, with a potential energy $U$

$$W=\Delta U+W_e=\Delta K$$ where $U$ is a potential energy ($q\Delta V$ for a charge in a potential field), $W_e$ is the work of other forces and/or non conservative forces and $\Delta K$ is the kinetic energy.

So, because on a potential energy surface $\Delta U =0$ then $\Delta K=W_e$ so yes, you move at constant speed if there is no extra force (if $W_e=0$ then $\Delta K =0$). If you want the particle to change speed (for example, if at the begininning your particle was not moving and you want to move it or, oppositely, if you want to stop it) you need to provide extra work $W_e=\Delta K$ but this work has to be in some other form: the electric (or gravitational) field can not help you (nor counter-act you!) on that surface!

So the statement "moving a charge on an equip. surface requires no work" is misleading and should rather be "the work done on a moving charge on an equipotential surface by the electric field is always 0" or "to change the state of motion of a charge in an equipotential field you need an external force" or "if you move a charge from one place to another on an equipotential surface the electric field has spent or gained 0 work, all the work that was spent/gained has been spent/gained by other fields".