# 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 $$r$$

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.

Note:
• 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 $$\displaystyle\frac{1}{r^3}$$.

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