Electric Field

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Section 11.2 Electric Field

Coulomb's law for the electric field (and the equivalent law for the gravitational field) at the point \(\rr\) due to a point charge \(Q\) (or point mass \(M\)) at the origin is

\begin{align*} \EE \amp= \frac{1}{4\pi\epsilon_0}\> \frac{Q\,\rhat}{r^2} ,\\ \gv \amp= -G\> \frac{M\,\rhat}{r^2} . \end{align*}

(Geometrically, what does the coordinate \(r\) represent here?) (Why is there a relative minus sign between the two fields?) This form of Coulomb's law most clearly shows the dimensions of the field (\(\rhat\) is dimensionless) and that the fall-off goes like \(r^{-2}\text{.}\) However, by realizing that \(\rr=r\,\rhat\text{,}\) we find an equivalent expression that is often easier to evaluate because it does not require us to normalize the unit vector \(\rhat\text{:}\)

\begin{align*} \EE \amp= \frac{1}{4\pi\epsilon_0} \frac{Q\,\rr}{|\rr|^3} ,\\ \gv \amp= -G \frac{M\,\rr}{|\rr|^3} . \end{align*}

These expressions could also be obtained by taking the (negative of the) gradient of the potentials \(V\) and \(\Phi\text{,}\) which are given by

\begin{align*} \EE = -\grad V \amp= -\grad\left(\frac{1}{4\pi\epsilon_0} \> \frac{Q}{r}\right) ,\\ \gv = -\grad \Phi \amp= \grad\left(G \> \frac{M}{r}\right) . \end{align*}

Masses and (positive) test charges will accelerate in the direction of the corresponding field, that is, in the direction in which the potential decreases.

Activity 11.2.1. Drawing the electric field.

Sketch the electric field for a single, positive charge located at the origin.

The electric field is a vector field in three dimensions. You may want to start by drawing this vector field in the \(xy\)-plane only, before considering points not in this plane.

The electric field due to a (positive) point charge at the origin is shown below.

Figure 11.2.1. The electric field of a (positive) point charge at the origin. (Click and drag the figure to see its three-dimensional nature.)