# Potential difference between two object

How am I to think about the potential difference between two objects? Non-point objects. I am reading a text that confuses me. It talks about the potential difference between two objects, yet it offers a calculation $$V=-\int^a_b \vec{E}\cdot d\vec{s}$$ of a line integral. A line (contour) goes between two points... It does not account for an entire object.

I thank you in advance, for any help. I am fluent in mathematics, and am now looking into physics out of interest.

• physics.stackexchange.com/questions/8221/… – pentane Aug 19 '18 at 17:56
• The text you cite speaks of metals, which we take to be ideal conductors. Every point inside a perfect conductor is at the same potential. So in the special case of objects made of a perfect conductor, you can define the potential difference between the objects as the potential difference between any point of the first to any point on the second – garyp Aug 19 '18 at 22:40

Potential differences are measured between two points in space, not between two objects.

If you want to find the potential energy of an object, you look at all the charges in the object and multiply them by the potential at the point in space where they are located.

• Thank you for your answer. When you say "look at all the charges in the object and multiply them by the potential at the point in space where they are located." does this not leave me with many values, one for each charge in an object? Perhaps I should also sum/integrate them over the object? – LeastSquaresWonderer Aug 19 '18 at 23:07
• Yes, that's exactly right, the potential energy is the sum of these values. – Ricky Tensor Aug 20 '18 at 5:53

The potential difference $V$ between two points is the work per unit charge required to move the charge between the points in an electric field. These may be between single points, between a point and a line of charge, or between two parallel sheets of charge, as well as many other configurations.

For example, for two parallel plates one positively charged and one negatively charged separated by distance $d$ with a uniform (constant) electric field $E$ between them (so E comes out of the integral), the magnitude of the potential difference $V$ between them is:

$$V =Ed$$

Hope this helps

• The work integral is always done between to points. You can't integrate from a "line of charge." One might integrate from a point anywhere on an equipotential surface. Don't confuse a charge collection with an equipotential. That line or sheet of charge might not be uniformly distributed. – Bill N Aug 19 '18 at 20:45
• @BillN Yes but it can be between two isolated charges, between a point charge and a uniform line of charge, between a point charge and a uniform sheet of charge, etc. – Bob D Aug 20 '18 at 9:46