Electric Fields

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

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 simulation a few times to push the electrons around.

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

q +1C E r

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

\(E\) = electric field at a point [N/C, N·C-1] vector
\(k_e\) = 8.987 × 109 = Coulomb's constant [N m²/C²] scalar
\(q\) = charge [C, Coulomb] scalar
\(r\) = distance between the charge and a location [m] scalar

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 at 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 a point [N/C, N·C-1] vector
\(q\) = charge [C, Coulomb] scalar

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 what force positive charge will experience. Our Charge is positive so it will go in the same direction as the field, right.

Example: What strength and direction of an electric field 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 to get a positive charge to go right

$$E = 2.087 \times 10^{-8} {\scriptsize \frac{N}{C}} \rightarrow$$

Solving for Multiple Charges

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.


Draw a picture

+1μC -2μC 50mm

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.


Draw a picture

-0.01μC 0.03μC 2m 1m

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

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