Electric field


My notes of the excellent lecture 2 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

If we have a charge positive \(Q\) and positive test charge \(q\), separated by distance \(r\)

Coulomb’s Law

The force \(\vec F\) on test charge \(q\) follows form Coulomb’s law $$ \vec F = \frac{q\,Q\,K}{r^2}\,\hat r \nonumber $$

The electric field \(\vec E\) at location \(P\) is defined as the force per unit charge $$ \shaded{ \vec{E_P} = \frac{\vec F}{q} = \frac{Q\,K}{r^2}\,\hat r } \quad \left[\frac{N}{C}\right] \tag{electric field} $$


  • by convention the test charge is positive,
  • the electric field tells you about the charge configuration.

Graphical representations

An electrical field is a vector field. The vectors point in the direction that a positive test charge would experience.

\(+3\) charge
\(-1\) charge

  • There are infinite number of vectors, but we only draw a handful
  • The length of the vector gives you an idea of the magnitude
  • Very close to the charge the arrows are larger, and off at one of \(r^2\). This makes it hard to see vectors further away from the charge.

Multiple charges

If we have multiple charges, \(q_1\ldots q_i\), then each charge will contribute to the electric field \(\vec E_P\) at point \(P\)

Superposition principle

Applying the superposition principle from Coulomb’s law $$ \shaded{ \begin{align*} \vec{E_P} &= \vec{E_1} + \vec{E_2} + \vec{E_3} + \cdots \\ &= \sum_i \vec{E_i} \end{align*} } \nonumber $$

If I place a charge \(q\) at location \(P\) in electric field \(\vec E\), then the force \(\vec F\) on that charge $$ \shaded{ \vec F = q\,\vec E } \nonumber $$


  • if \(q\) is negative, the force will be in the opposite direction of \(\vec E\),
  • if \(q\) is large the force will be large
  • if \(q\) is small the force will be small


Suppose, two charges: \(q_1=+3\), and \(q_2=-1\)

Charges \(+3\) and \(-1\)


  • Because of the with inverse \(r^2\) relationship, just to the right of \(-1\) that charge will win. Far to the right, the distance between the charges doesn’t mater, the field will be as if there was a \(+3-1=+2\). Somewhere in between, there is a point \(p\) where the charges cancel each other out, and \(\vec E=\vec 0\).
  • Anywhere far away from the charges, the vectors are pointing out. There, the configuration behaves as \(+3-1=+2\).

Field lines

A positive test particle placed on a field line, will experience a force tangential with that field line. For same charge configuration

Field lines from \(+3\) and \(-1\) charges


  • The field lines tell you in what direction a positive charge will experience a force.
  • There are infinite field lines (as there are vectors)
  • You can think about the field lines as air moving from a source to a sink.
  • With field lines we lost the information about the magnitude of the field. But where the density of the lines is high the field is stronger.
  • You can visualize field lines using grass seeds suspended in oil.


Field lines are not trajectories.

\(q\) in parallel field \(q\) in non-parallel field
If you release a charge with \(0\) speed, it will accelerate along the field line and stay on the field line. If you release a charge with \(0\) speed, it will accelerate tangential to the field line, but immediate abandon that field line.


The special case where the charges are equal in magnitude, but opposite in sign

Dipole field lines

Let \(Q\) and \(-Q\) be point charges, placed at \((-0.5,0)\) and ((0.5,0)\). The field on the \(x\)-axis at \(x=r\) for \(r\gt 0\) for the charges $$ \begin{align*} E_+ &= \frac{K(+Q))}{\left(r-0.5\right)^2},\quad E_- = \frac{K(-Q))}{\left(r+0.5\right)^2} \nonumber \\ \Rightarrow E &= E_+ + E_- = \frac{KQ}{r^2\left(1-\tfrac{1}{2r}\right)^2} – \frac{KQ}{r^2\left(1+\tfrac{1}{2r}\right)^2} \end{align*} $$

Since \(\frac{1}{2r}\ll 1\), approximate using binominal series expansion $$ \frac{1}{\left(1\pm \tfrac{1}{2r}\right)^2} \approx 1\mp \frac{1}{r} \nonumber $$

Substitute the expansion $$ \begin{align*} E &= \frac{KQ}{r^2}\left( \left(\bcancel{1}+\tfrac{1}{r}\right) – \left(\bcancel{1}-\tfrac{1}{r}\right) \right) \\ &= \frac{2KQ}{r^3} \end{align*} $$

So, if you are far away from a dipole, the field falls off at \(\displaystyle\frac{1}{r^3}\).

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