Students will learn what an electric field is, how to draw electric field lines and how to calculate electric fields. They will also learn to apply the electric field in order to find the direction a charge would move upon entering the field and the force on it.

### Key Equations

\begin{align*}E = \frac{k q}{r^2} \; \; \end{align*} ; Electric field due to charge \begin{align*}q\end{align*} at a distance *r* from the charge.

\begin{align*}F = q E \; \; \end{align*} ; Force due to an electric field.

### Guidance

Gravity and the Coulomb force have a nice property in common: they can be represented by **fields**. Fields are a kind of bookkeeping tool used to keep track of forces. Take the electromagnetic force between two charges given above: \begin{align*}\vec{F_e} = \frac{kq_1q_2}{r^2}\end{align*} If we are interested in the acceleration of the first charge only --- due to the force from the second charge --- we can rewrite this force as the product of \begin{align*}{q_1} \end{align*} and \begin{align*}\frac{kq_2}{r^2}\end{align*}. The first part of this product only depends on properties of the object we're interested in (the first charge), and the second part can be thought of as a property of the point in space where that object is.

In fact, the quantity \begin{align*}\frac{kq_2}{r^2}\end{align*} captures everything about the electromagnetic force on any object possible at a distance **r** from \begin{align*} q_2 \end{align*}. If we had replaced \begin{align*} q_1 \end{align*} with a different charge, \begin{align*} q_3 \end{align*}, we would simply multiply \begin{align*} q_3 \end{align*} by \begin{align*}\frac{kq_2}{r^2}\end{align*} to find the new force on the new charge. Such a quantity, \begin{align*}\frac{kq_2}{r^2}\end{align*} here, is referred to as the electric field from charge \begin{align*} q_2 \end{align*} at that point: in this case, it is the electric field due to a single charge: \begin{align*}
\vec{E_f} = \frac{kq}{r^2}
\end{align*}

The electric field is a vector quantity, and points in the direction that a force felt by a positive charge at that point would. If we are given the electric field at some point, it is just a matter of multiplication --- as illustrated above --- to find the force any charge \begin{align*} q_0 \end{align*} would feel at that point: \begin{align*}
\underbrace{\vec{F_e}}_{\text{Force on charge }q_0} = \underbrace{\vec{E_f}}_{\text{Field}} \underbrace{\times q_0}_{\text{Charge}} &&\text{Force on charge }q_0\text{ in an electric field}
\end{align*} Note that this is true for *all* electric fields, not just those from point charges. In general, the **electric field** at a point *is the force a* *positive test charge of magnitude 1**would feel at that point*. Any other charge will feel a force along the same line (but possibly in the other direction) in proportion to its magnitude. In other words, the electric field can be thought of as "force per unit charge".

In the case given above, the field was due to a single charge. Such a field is shown in the figure below. Notice that this a field due to a positive charge, since the field arrows are pointing outward. The field produced by a point charge will be radially symmetric i.e., the strength of the field only depends on the distance, \begin{align*} r \end{align*}, from the charge, not the direction; the lengths of the arrows represent the strength of the field.

#### Example 1

**Question**: Calculate the electric field a distance of \begin{align*}4.0\mathrm{mm}\end{align*} away from a \begin{align*}-2.0\mu \mathrm{C}\end{align*} charge. Then, calculate the force on a \begin{align*}-8.0\mu \mathrm{C}\end{align*} charge placed at this point.

**Answer**: To calculate the electric field we will use the equation \begin{align*}
E=\frac{kq}{r^2}
\end{align*} Before we solve for the electric field by plugging in the values, we convert all of the values to the same units.

\begin{align*} 4.0\mathrm{mm}\times \frac{1\mathrm{m}}{1000\mathrm{mm}}&=.004\mathrm{m}\\ -2.0\mu \mathrm{C}\times \frac{1\mathrm{C}}{1000000\mu \mathrm{C}}&=-2.0\times 10^{-6}\mathrm{C} \end{align*}

Now that we have consistent units we can solve the problem.

\begin{align*} E=\frac{kq}{r^2}=\frac{9\times 10^9\mathrm{Nm^2/C^2}\times -2.0\times 10^{-6}\mathrm{C}}{(.004\mathrm{m})^2}=-1.1\times 10^9\mathrm{N/C} \end{align*}

To solve for the force at the point we will use the equation \begin{align*} F=Eq \end{align*} . We already know all of the values so all we have to do is convert all of the values to the same units and then plug in the values.

\begin{align*} -8.0\mu \mathrm{C}\times \frac{1\mathrm{C}}{1000000\mu \mathrm{C}}=-8.0\times 10^{-6}\mathrm{C}\\ \end{align*}

\begin{align*} F=Eq=-8.0\times 10^{-6}\mathrm{C} \times -1.1 \times 10^9 \mathrm{N/C}=9000\mathrm{N} \end{align*}

### Watch this Explanation

### Simulations

Electric Field of Dreams (PhET Simulation)

### Explore More

- What is the direction of the electric field if an electron initially at rest begins to move in the North direction as a result of the field?
- North
- East
- West
- South
- Not enough information

- Two metal plates have gained excess electrons in differing amounts through the application of rabbit fur. The arrows indicate the direction of the electric field which has resulted. Three electric potential lines, labeled \begin{align*}A, B,\end{align*} and \begin{align*}C\end{align*} are shown. Order them from the greatest electric potential to the least.
- \begin{align*}A, B, C\end{align*}
- \begin{align*}C, B, A\end{align*}
- \begin{align*}B, A, C\end{align*}
- \begin{align*}B, C, A\end{align*}
- \begin{align*}A = B = C \ldots\end{align*} they’re all at the same potential

- The three arrows shown here represent the magnitudes of the electric field and the directions at the tail end of each arrow. Consider the distribution of charges which would lead to this arrangement of electric fields. Which of the following is most likely to be the case here?
- A positive charge is located at point \begin{align*}A\end{align*}
- A negative charge is located at point \begin{align*}B\end{align*}
- A positive charge is located at point \begin{align*}B\end{align*} and a negative charge is located at point \begin{align*}C\end{align*}
- A positive charge is located at point \begin{align*}A\end{align*} and a negative charge is located at point \begin{align*}C\end{align*}
- Both answers a) and b) are possible

- Particles \begin{align*}A\end{align*} and \begin{align*}B\end{align*} are both positively charged. The arrows shown indicate the direction of the
*forces*acting on them due to an applied electric field (not shown in the picture). For each, draw in the electric field lines that would best match the observed force. - Calculate the electric field a distance of \begin{align*}4.0 \;\mathrm{mm}\end{align*} away from a \begin{align*}-2.0 \ \mu \mathrm{C}\end{align*} charge.
- Copy the arrangement of charges below. Draw the electric field from the \begin{align*}-2 \;\mathrm{C}\end{align*} charge in one color and the electric field from the \begin{align*}+2 \;\mathrm{C}\end{align*} charge in a different color. Be sure to indicate the directions with arrows. Now take the individual electric field vectors, add them together, and draw the resultant vector. This is the electric field created by the two charges together.
- A proton traveling to the right moves in between the two large plates. A vertical electric field, pointing downwards with magnitude \begin{align*}3.0 \;\mathrm{N/C}\end{align*}, is produced by the plates.
- What is the direction of the force on the proton?
- Draw the electric field lines on the diagram.
- If the electric field is \begin{align*}3.0 \;\mathrm{N/C}\end{align*}, what is the acceleration of the proton in the region of the plates?
- Pretend the force of gravity doesn’t exist; then sketch the path of the proton.
- We take this whole setup to another planet. If the proton travels straight through the apparatus without deflecting, what is the acceleration of gravity on this planet?

- A molecule shown by the square object shown below contains an excess of \begin{align*}100\end{align*} electrons.
- What is the direction of the electric field at point A, \begin{align*}2.0 \times 10^{-9} \;\mathrm{m}\end{align*} away?
- What is the value of the electric field at point \begin{align*}A\end{align*}?
- A molecule of charge \begin{align*}8.0 \ \mu \mathrm{C}\end{align*} is placed at point \begin{align*}A\end{align*}. What are the magnitude and direction of the force acting on this molecule?

**Answers**

- .
- .
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- \begin{align*}1.1 \times 10^9 \;\mathrm{N/C}\end{align*}
- .
- a. down b. Up \begin{align*}16c, 5.5 \times 10^{11} \;\mathrm{m/s}^2\end{align*} e. \begin{align*}2.9 \times 10^8 \;\mathrm{m/s}^2\end{align*}
- a. Toward the object b. \begin{align*}3.6 \times 10^4 \;\mathrm{N/C}\end{align*} to the left c. \begin{align*}2.8 \times 10^{-7} \;\mathrm{N}\end{align*}