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[NOTES/EM-03021] Intuitive Proof of Path Independence of Work in Electrostatics

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An intuitive proof of path independence of work done by electrostatic forces is given following Feynman.

We now give an intuitive argument that the work done by the electric field of point charge located at the origin is given by \eqref{PathInd} An arbitrary path, such as a path \(AA_1A_2A_3B\) shown in \Figref{ArbPath}, can be approximated by paths, \(\gamma_1, \gamma_2, \gamma_3,\gamma_4\) consisting of circular arcs and line segments as in special paths I,I,III discussed above. For these paths the work done by the electric field of a point charge at the origin, is

  \begin{eqnarray}
 \int_{\gamma_1}\vec E \cdot \overrightarrow{dl} &=& \frac{1}{r_A} - \frac{1}{r_1}\\  
   \int_{\gamma_2}\vec E \cdot \overrightarrow{dl} &=& \frac{1}{r_1} - \frac{1}{r_2}\\
  \int_{\gamma_3}\vec E \cdot \overrightarrow{dl} &=& \frac{1}{r_2} - \frac{1}{r_3}\\
  \int_{\gamma_4}\vec E \cdot \overrightarrow{dl} &=& \frac{1}{r_3} - \frac{1}{r_B}.
\end{eqnarray}

 

  

The sum of the four contributions to the work \begin{equation} \sum_k \int_{\gamma_k}\vec E \cdot \overrightarrow{dl} = \frac{1}{r_A} - \frac{1}{r_B}. \end{equation} depends only on the end points \(A,B\).

Thus, by making a larger number of subdivisions of the \(AB\), we see that the line integral along the full path remains a function of the end points only. Since this path \(AB\) is arbitrary the  path independence of work done by the electric field will hold for all  paths

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