I was recently solving a question in which I was given a large object in a uniform electric field. I was able to solve it by assuming that the force acts from the centre of mass of the object drawing up an analogy to gravity.

I am also aware of the fact that centre of mass and centre of gravity coincide only when the gravitational field is uniform. I feel the same should be true here.

However, I have never heard about "Centre of Electric Field".

So where does the Electric Field act on a large body in the most general case?

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    $\begingroup$ That would be the "centre of charge". Charge may be distributed differently than mass, though, and the distribution of charge may even be influenced by the external field $\endgroup$ Dec 14, 2020 at 9:30
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    $\begingroup$ Related: physics.stackexchange.com/q/8221 $\endgroup$ Dec 14, 2020 at 10:44

2 Answers 2


Suppose you have a macroscopic body that's charged. We can assume that the charge will be distributed in some way into the body. In this situation, by definition, the total electric force acting on the body is the sum of all the little forces that the charged point like subsection of the body experience, but if we want to thing about the charged body as made of point like subsection then we get an infinite amount of infinitesimal subsection, this implies that we have to perform the integral instead of the sum.

Long story short: the electric force experienced by a single point like subsection of the body is: $$d\vec{F}=\vec{E}\rho dV$$ where $\vec{E}$ is of course the electric field in that point and $\rho$ is the charge density.
So to get the total force $\vec{F}$ acting on the body we have to perform the integral: $$\vec{F}=\int _V \rho(\vec{x}) \vec{E}(\vec{x})dV$$ This is it. However: A fundamental law of nature is that nobody wants to compute integrals! Can we find some handy shortcut? Well: if the electric field $\vec{E}$ is constant in space (or if we can approximate it this way ;)) there is a pretty nice simplification: $$\vec{F}=\int _V \rho(\vec{x}) \vec{E}(\vec{x})dV=\vec{E}\int _V \rho(\vec{x})dV=\vec{E}Q$$ where $Q$ is the total charge of the body, an analogous concept to the total mass, but of course not the same thing. Of course in this case the electric force is applied in the center of charge ($\vec{C}$), analogous concept to the center of mass: $$\vec{C}=\frac{1}{Q}\int _V\rho(\vec{x})\vec{x}dV$$ Even better: if the charge distribution is simmetrically distributed in some nice way we can easily guess the position of $\vec{C}$ most of the times, this way we don't have to perform the integral!

Keep in mind: in some situation the center of mass and the center of charge share position, but this does not happen most of the times! In fact to calculate the position of the center of mass you integrate the mass density, but to calculate the position of the center of charge, as we saw, you have to integrate the charge density.


Assuming a rigid body i.e: a body that doesn't deform then no matter what reason the forces occurred, we can always say the whole body can be said to be a point particle whose motion is governed by the net forces acting on that body.

And about 'where the electric field' acts.. it acts everywhere in a way. Whenever you have a few charges causing an electric field, then the field is defined for each and every point in space. To find the net field at a point we have to do the vector sum of fields.

To find the net force exerted on a body by an external field, we need to dot integrate the product of charge density produced with the field over the whole 'measure' of the body.

By I measure I mean area if 2-D surface, volume if 3-D etc.


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