**Forces** come in two flavors: **contact** and **non-contact**. The contact forces are simple. If I push a cart with my hand, I have to *touch* it to do so. I can't make it move by waving my hands at it.

On the other hand (no pun intended), several forces are exerted "at a distance." **Gravity**, the **electrostatic force** (the force between charges) and the **nuclear forces** are examples of such invisible forces.

**Electric fields** are a result of the electrostatic force, sometimes referred to as the **Coulomb force**, for **Coulomb's law**:

$$F_{es} = \frac{k q_1 q_2}{r^2}$$

where **k** is the **Coulomb constant** $\left( k = 8.99 \times 10^9 \, \frac{N m^2}{C^2}\right),$ **q _{1}** and

The force exerted on any charge by another set of electric charges (or just one charge) is proportional to the amounts of charge and inversely-proportional to the square(s) of the distance(s) between them.

Typically, we employ an imaginary charge, a positive "**test charge**" to map out or probe the forces around another group of charges. That map will be what we'll call the electric field.

The electric field is a vector field, or a set of vectors that give the strength and direction of the force that our test charge would "feel" at any point near another group of charges.

In the examples below, we'll map out a few simple electric fields so you can see how this works. We'll start with the fields around single positive and negative charges.

**test charge**" to probe the strength and directions of electric fields. By long-standing convention, the test charge is **positive** and has a charge of +1. The test charge helps us to visualize how a field can be mapped according to the relative forces at a given position in the field.

The diagram below shows a positive charge with four test charges ( ) placed at no particular position. The vector arrows represent the repulsive force that a positive test charge would feel due to the presence of a positive charge.

Notice that the closer the test charge, the greater the repulsive force, and thus the longer the force vector. If we're a little more systematic, we can place test charges at regular intervals and come up with a force-vector diagram like this:

Of course, this is a 2D representation of a 3-dimensional concept, but you get the idea. Finally, this force field is often represented as we might represent land contours on a contour map, like this:

In such a field diagram, the forces, and thus the electric-field strength, are greater where the circles (contours) are closer together, and weaker where they're farther apart.

We can sketch analogous diagrams for the electric field of a point negative charge. The only difference is that the forces are attractive. Remember, the test charge is always positive by convention. Here are the force vectors for a few test-charge positions:

This time the vectors point toward the negative charge because the force is attractive. The test charge is placed in enough locations for us to get a good feel for what the force field or electric field will look like.

The contour diagram looks the same: strong attractive forces near the negative charge, diminishing with the square of the distance as we move away from it.

In the sections that follow, we'll look at a couple more distributions of charges and see what their electric fields look like.

The electric fields described in this section are **electrostatic fields**. "Static" means not moving. They are produced by sets of charges that are fixed in space relative to each other. Fields produced by moving charges ("dynamic" = in motion) are more difficult to describe mathematically.

The **electric dipole** can be viewed simply as two charged particles separated by a distance, or as two "**poles**" with some separation. The latter definition will allow us to expand the idea of the dipole to objects, like molecules, that are made of more than one particle (e.g. atom), but still have a negative end and a positive end.

Here is a typical dipole electric field diagram. This dipole is composed of a positive and a negative charge. Three test charges are located along one of the field lines.

The vectors signifying repulsion from the positive pole and attraction to the negative pole and the total force vector have this color scheme:

The field lines are built up by moving the test charge around, say on a fine grid, and calculating the net force on it at each point. Those results are then displayed as a set of smooth lines.

The closer the lines in the electric field diagram, the more electrostatic force (Coulomb's law) would be felt be a test charge. The density of lines is thickest near each charge, and we'd expect the highest forces there.

The electric dipole is very important in physics and chemistry. This diagram shows an inert gas atom, which is more-or-less a sphere with a symmetric charge distribution – positive in the center (nucleus) surrounded by negatively-charged electrons. There's no + and - end of an argon or neon atom.

On the other hand, HCl consists of an atom with a lot of electrons bound to an atom with one electron. Further, the Cl atom is good at drawing the lone **e ^{-}** of the hydrogen toward itself. It clearly has a

Water, which is made of *three* atoms also behaves as a dipole. The hydrogen atoms are more positively charged than the lone pairs (contours in the diagram) of electrons on the opposite side of the oxygen. Water actually has quite a strong dipole for its size. Larger molecules composed of many more atoms can also act as dipoles. All that's required is that there is an imbalance in the charge distribution.

Two like charges can form a **repulsive dipole**. This diagram shows two positive charges located at some fixed distance from each other.

Four test charges are randomly located in the field so that you can see how the repulsive force vectors due to each positive charge on the test charge sum to a vector aligned with the curved lines of the field.

The red dashed lines are there to help you see where the directions of the various repulsive vectors come from.

We could form a similar repulsive dipole using two negative charges. The arrows of the field would just have to be reversed.

Here's an arrangement of electric charges called a quadrupole – four poles. These are arranged at the corners of a square.

Several test charges are shown. Each is under the influence of each of the four fixed charges. The red vectors are repulsive, the black vectors attractive and the blue vectors are tangent to what will eventually be our field lines once we test enough points.

Again, this is a 2D representation of what would be a 3D field, so keep that in mind.

The completed field diagram will look something like this.

The quadrupole field has a lot of symmetry, and quad fields, both magnetic electric quadrupole fields, are useful in steering beams of charged particles like electrons in many devices.

The electric field is a force field that depends upon the potential energy difference between two or more physical points, and upon the charges involved.

The units of electric field are force divided by charge. The SI units are **Newtons per Coulomb**:

$$\frac{N}{C} = \frac{Kg\cdot m}{s^2 \cdot C}$$

Another common way to express electric field strength is **Volts per meter**. Volts are the unit of electric potential, energy divided by charge.

$$ \begin{align} \frac{V}{m} &= \frac{\text{Potential energy}}{\text{Coulomb·m}} \\ \\ &= \frac{J}{C\cdot m} = \frac{Kg\cdot m^2}{s^2\cdot C \cdot m} \end{align}$$

You can see that the basic SI units are the same.

The SI units of electric charge are Newtons per Coulomb (N/C) or Volts per meter (V/m).

X
### SI units

SI stands for Système international (of units). In 1960, the SI system of units was published as a guide to the preferred units to use for a variety of quantities. Here are some common SI units

length | meter | (m) |

mass | Kilogram | (Kg) |

time | second | (s) |

force | Newton | N |

energy | Joule | J |

Calculate the electric field produced by a single AA-sized battery.

**Solution**

$$E = \frac{1.5 \; V}{0.05 \; m} = 30 \; \frac{V}{m}$$

Calculate the size (magnitude) of an electric charge that would create an electric field of 1.0 N/C at a point 1 meter away.

**Solution**

$$E = \frac{k q}{r^2} \; \longrightarrow \; q = \frac{r^2 E}{k}$$

Plugging what we know into the right side and canceling units gives us our charge:

$$ \begin{align} q &= \frac{(1 m)^2 \cdot 1 NC^{-1}}{9 \times 10^9 \, Nm^2C^{-2}} \\ \\ &= 1.11 \times 10^{-10} \; C \end{align}$$

Four charges are arranged as shown in the diagram below. Calculate the direction and magnitude of the force that would be felt by a test charge located at the center ( **X** ).

**Solution****F _{1} – F_{4}**. Each has components along the horizontal and vertical axes, as the figure is drawn.

Now those four force vectors are going to add to give us our net force. Here's the graphical vector addition picture, just so we know what our calculation should yield:

Now each vector can be resolved as a sum of vectors in the horizontal and vertical directions,

The labeling is changed because we'll ignore our test charge (+1 C) in the calculations. These will now be field vectors, not strictly force vectors, but numerically, they'll be the same.

A little right triangle geometry gives us the distance from the test charge to any of the four quadrupole charges as **r** = 3.5636 cm = 0.035636 m (as usual, we'll keep a lot of digits around until the end of the calculation to avoid cumulative round-off errors)

Now the magnitudes of the field vectors between the test charge and the larger charges is:

$$ \begin{align} E_1 = E_2 &= \frac{kq}{r^2} \\ \\ &= \frac{9 \times 10^9 Nm^2C^{-2} (2.0 \times 10^{-8} C)}{(0.03536 \, m)^2} \\ \\ &= 143,962 \; N/C \end{align}$$

Using the right triangle relationships gives the lengths of **E _{1}** and

$$|E_{1x}| = |E_{1y}| = 101,796 \; N/C$$

It is apparent from the diagram that these point in opposite **x**-directions, so they'll cancel in that dimension (but they'll add in the **y**-direction).

We can follow the same procedure for finding the **x**-components of the field vectors between the test charge and the smaller charges – first the magnitudes of **E _{3}** and

$$ \begin{align} E_3 = E_4 &= \frac{kq}{r^2} \\ \\ &= \frac{9 \times 10^9 Nm^2C^{-2} (1.0 \times 10^{-8} C)}{(0.03536 \, m)^2} \\ \\ &= 71,981 \; N/C \end{align}$$

And the magnitudes of the **x**-components of those are then

$$|E_x| = |E_y| = 50,898 \; N/C$$

These components also face in opposite directions in the **x**-dimension, so the sum of all **x**-components is zero:

$$ \begin{align} |E_x| = 50,898 &- 50,898 \\ &+ 101,796 + 101,796 = 0 \end{align}$$

The **y**-components of all four vectors, because of the symmetry all have the same lengths or magnitudes as the **x**-components, but they have different directions. The sum of the **y**-components is:

$$ \begin{align} |E_y| = 2(-50,898) &+ 2(101,796) \\ &= 101,796 \; N/C \end{align}$$

where the absolute value bars in this case mean "length" or "magnitude."

So the net electric field felt by our test charge is **101,796 N/C** in the "up" or **+y** direction.

1. |
Calculate the magnitude (size) of a point charge that would create an electric field of 1.50 N/C at a distance of 1 m. |

2. |
The figure below shows two point charges (q |

3. |
Two particles with equal charge of 2.4 × 10 |

4. |
Two charges, equal in magnitude (1.0 × 10 |

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