Electric Potential Energy and Electric Potential

Imagine a positive charge q that is placed in a region that has a constant electric field, E, pointing to the left. The charge would experience a force, qE, also pointing to the left.

We would like to slowly move the particle from position x₁ to position x₂ by applying an external force to the right that is slightly bigger than qE. Therefore, we let F_ext= eE. (See the figure below.)

← E

___|__________|_____________________|___________

x=0 x₁ x₂ +x-axis

By definition, the external work would equal,

W_ext º F_ext Dx cosq = F_ext Dx cos0 = qE Dx , where Dx=x₂ – x₁ .

The increase in electrical potential energy of q is defined to equal the work,

DEPE º W_ext

The change in electric potential between points 1 and 2 is defined,

DV = V₂ – V₁ º DEPE/q = W_ext/q

Therefore,

DV = E Dx (constant E).

If instead, E was that produced by a point charge Q then calculus shows,

DV = kQ( 1/r₂ – 1/r₁) (point charge Q)

where I have made the substitution x→r. Often we assume r₁ → ∞ and take V(∞) = 0,

we say ground is at infinity and so,

V(r) = kQ/r (point charge Q).

So, surrounding a positive point charge Q there is an electric field pointing away from Q, and a region where the voltage is positive. Surrounding a negative charge, the field points toward the charge and the voltage is negative.

Examples[in class]