##### Electrostatic Field of Point Charges
ems.es.ef.pc

Our tendency is to think that force acts locally, not remotely. It is thus natural to imagine a "vehicle" through which one charge exerts a force on another. This leads us to an important first abstraction, that of the electrostatic field.

Going back to our source and test charge setup, we shall thus think of the force on the test charge as being given by its coupling to an electric field originating from the source charge. Invoking the superposition principle, we can thus write

- Gr (2.3)

\[ {\bf F}_t = q_t {\bf E}({\bf r}_t) \tag{FqE}\label{FqE} \]

with the electric field of a point charge distribution being

- Gr (2.4)

\[ {\bf E} ({\bf r}) \equiv \frac{1}{4\pi \varepsilon_0} \sum_{i=1}^n q_i \frac{{\bf r} - {\bf r}_i}{|{\bf r} - {\bf r}_i|^3} \tag{E_pcd}\label{E_pcd} \]

The electric field \({\bf E} ({\bf r})\) is thus the force per unit charge that would be exerted if you put a test charge at position \({\bf r}\).

Example

Find the field at distance \(z\) above the midpoint between two equal charges \(q\) placed a distance \(d\) apart from each other.

Solution: by symmetry, only the \(\hat{\bf z}\) part of the field is nonvanishing. We have \(|{\bf r} - {\bf r}'| = \sqrt{z^2 + (d/2)^2}\) for both charges and \(\cos \theta = z/|{\bf r} - {\bf r}'|\) so

\[ {\bf E} = \frac{1}{4\pi \varepsilon_0} \frac{2q z}{[z^2 + d^2/4]^{3/2}} ~\hat{\bf z} \]

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42