A
**field** has a value for each point in space and time. For example, on a weather map, the
surface wind velocity is described by assigning a vector to each point on the map.

Electric fields predict the electric force of an imaginary +1C test charge at a location. This is useful when you only know about one charge instead of the pair, but remember the test charge doesn't exist.

Click the simulations to push the electrons out of position.

Why does the field go to zero when the protons and electrons are in atoms?

## Electric Field Equation

We get the electric field equation by simply setting one q to +1C in the electrostatic force equation.

## $$ E = \frac{k_{e}q}{r^{2}}$$

\(E\) = electric field at a point [N/C, N·C
^{-1}]
*vector*

\(k_e\) = 8.987 × 10
^{9} = Coulomb's constant [N m²/C²]

\(q\) = charge [C, Coulomb]

\(r\) = distance between the charge and a location [m]

We can figure out the direction of the electric field by imagining what a positive +1C charge would do at each location.

Negative charges produce an inward electric field.

Positive charges produce an outward electric field.

**Example:**Find the electric field 2.0m away from a negative 3.0μC charge.

## solution

$$\mu = micro = 10^{-6}$$$$E = \frac{k_{e}q}{r^{2}}$$ $$E = \frac{(8.987 \times 10^{9})(-3 \times 10^{-6})}{(2)^{2}}$$ $$E = -6740 {\scriptsize \frac{N}{C}}$$

The test charge is always positive and the charge that generates this field is negative. Opposites attract, so the field is pointed towards the -3.0μC charge

**Example:**How far from a 10nC charge is the field strength 10N/C?

## solution

$$n = nano = 10^{-9}$$$$E = \frac{k_{e}q}{r^{2}}$$ $$r^{2} = \frac{k_{e}q}{E}$$ $$r^{2} = \frac{(8.987 \times 10^{9})(10 \times 10^{-9})}{10}$$ $$r^{2} = 8.987$$ $$r = 2.998m$$

## Finding Force in an Electric Field

Knowing the direction and strength of an electric field makes it easy to find the force felt by a charge placed in the field.

## $$F = Eq$$

\(F\) = electrostatic force [N, Newton, kg m/s²]
*vector*

\(E\) = electric field at the charge [N/C, N·C
^{-1}]
*vector*

\(q\) = charge [C, Coulomb]

The charge q is not the source of the electric field E.

**Example:**A positive 0.005μC charge is placed in a uniform 200N/C field pointed to the right. What force does the charge feel?

## solution

$$F = Eq$$ $$F = (200)(0.005 \times 10^{-6})$$ $$F = 10^{-6} N \ \rightarrow$$ The electric field shows the force a positive charge would experience. Our charge is positive so it will go in the same direction as the field,**right**.

**Example:**Find the strength and direction for an electric field that would make a proton accelerate to the right at 2m/s²?

## solution

$$m_{p} = 1.672 \times 10^{-27}kg \quad q_{p} = 1.602 \times 10^{-19}C$$$$F=ma$$ $$F=(1.672 \times 10^{-27})(2)$$ $$F=3.344 \times 10^{-27} N$$

$$F = Eq$$ $$\frac{F}{q} = E$$ $$\frac{3.344 \times 10^{-27}N}{1.602 \times 10^{-19}C} = E$$ $$2.087 \times 10^{-8} {\scriptsize \frac{N}{C}} = E$$

The field is directed right so a positive charge will go right.

$$E = 2.087 \times 10^{-8} {\scriptsize \frac{N}{C}} \rightarrow$$## Solving for Multiple Charges

To find the field from multiple charges, we need to add each individual field contribution from each charge. Be mindful of the direction of each field contribution to see which are negative or positive.

**Example:**A positive 1μC charge is 50mm away from a -2μC charge. Find the magnitude and direction of the electric field halfway in between these charges.

## solution

Draw a picture

Find the electric field from each charge at the location.

$$\mu = 10^{-6}$$ $$E = \frac{k_{e}q}{r^{2}}$$ $$E = \frac{(8.987 \times 10^{9})(10^{-6})}{(0.025)^{2}}$$ $$E = 14,379,200 {\scriptsize \frac{N}{C}} \rightarrow$$$$E = \frac{(8.987 \times 10^{9})(-2 \times 10^{-6})}{(0.025)^{2}}$$ $$E = 28,758,400 {\scriptsize \frac{N}{C}} \rightarrow$$

Add the fields and keep in mind direction

$$14,379,200 + 28,758,400 = 43,137,600 {\scriptsize \frac{N}{C}} \rightarrow $$**Example:**A 0.03μC charge is 2m East of a -0.01μC charge. Find the magnitude and direction of the electric field 1m West of the -0.01μC charge.

## solution

Draw a picture

Find the electric field from each charge at the location.

$$E = \frac{k_{e}q}{r^{2}}$$ $$E = \frac{(8.987 \times 10^{9})(-0.01 \times 10^{-6})}{(1)^{2}}$$ $$E = 89.87 {\scriptsize \frac{N}{C}} \rightarrow$$$$E = \frac{(8.987 \times 10^{9})(0.03 \times 10^{-6})}{(3)^{2}}$$ $$E = 29.95\overline{6} {\scriptsize \frac{N}{C}} \leftarrow$$

Add the fields and keep in mind direction

$$89.87-29.95\overline{6} =59.91\overline{3} {\scriptsize \frac{N}{C}} \rightarrow $$