\(\)
If we have a charge positive \(Q\) and positive test charge \(q\), separated by distance \(r\)
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 $$ \shaded{ \vec{E_P} = \frac{\vec F}{q} = \frac{Q\,K}{r^2}\,\hat r } \quad \left[\frac{N}{C}\right] \tag{electric field} $$
Notes:
- 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.
- 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 \(P\)
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 $$ \shaded{ \vec F = q\,\vec E } \nonumber $$
Note:
- 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
Quantitative
Suppose, two charges: \(q_1=+3\), and \(q_2=-1\)
Note
- 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
Note
- 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.
Trajectories
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. |
Dipole
The special case where the charges are equal in magnitude, but opposite in sign
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 \(\displaystyle\frac{1}{r^3}\).