Superposition for the Electric Field

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Section 11.3 Superposition for the 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?) This form of Coulomb's law most clearly shows the dimensions of the electric field (\(\rhat\) is dimensionless) and that the fall-off goes like \(\frac{1}{r^2}\text{.}\) 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*}

We define the electric and gravitational fields as

\begin{align*} \gv \amp = -\grad \Phi ,\\ \EE \amp = -\grad V . \end{align*}

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

These expressions could also be obtained by differentiating \(V\text{,}\) which is given by

\begin{gather*} V = \frac{1}{4\pi\epsilon_0} \> \frac{Q}{r} = \frac{1}{4\pi\epsilon_0} \> \frac{Q}{|\rr|} . \end{gather*}