Electric Charge

Lesson Plan

• Introduction to Electric Charge and its properties.
• Understanding Conductors and Insulators.
• Methods of Charging (Friction, Conduction, Induction).
• Exploring Coulomb's Law and its applications.
• Introduction to Electric Fields.
• Practice Problems (Simple, Intermediate, Hard).
• Interactive Activities & Discussion.
• Q&A and Summary.

Learning Objectives

• Define electric charge and identify its types.
• Describe the properties of electric charge.
• Differentiate between conductors and insulators.
• Explain and demonstrate the three methods of charging.
• State and apply Coulomb’s Law.
• Define and draw electric fields for point charges and parallel plates.
• Solve problems involving electric charge and fields.

Properties of Electric Charge

• Quantization of Charge: Charge exists in discrete units, a multiple of the elementary charge, $e$. The formula is $Q = ne$, where $n$ is an integer.
• Conservation of Charge: The net electric charge of an isolated system remains constant.
• Like charges repel, unlike charges attract.

Conceptual Problem

A student rubs a plastic rod with a piece of cloth. The rod becomes negatively charged. What happened to the cloth?

Solution

Since the plastic rod gained electrons to become negatively charged, the cloth must have lost electrons. This means the cloth is now positively charged.

Activity: Static Cling!

Take a balloon and rub it on your hair. Then, place the balloon next to a wall or a small piece of paper. What do you observe? Discuss with a partner why this happens.

Coulomb's Law

Coulomb's Law describes the electrostatic force $\vec{F}$ between two stationary, electrically charged particles.

The formula is:

$$\vec{F} = k \frac{q_1 q_2}{r^3}\vec{r}$$

Where:

• $\vec{F}$ is the electrostatic force vector.
• $q_1$ and $q_2$ are the magnitudes of the charges.
• $r$ is the distance between the charges.
• $\vec{r}$ is the position vector from $q_1$ to $q_2$.
• $k$ is Coulomb's constant, $8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2$.

Simple Problem

Two point charges, $q_1 = +2.0 \times 10^{-6} \, \text{C}$ and $q_2 = +3.0 \times 10^{-6} \, \text{C}$, are separated by a distance of $0.50 \, \text{m}$. Calculate the electrostatic force between them.

Solution

Using Coulomb's Law, the magnitude of the force is $F = k \frac{|q_1 q_2|}{r^2}$:

$$F = (8.99 \times 10^9) \frac{|(2.0 \times 10^{-6})(3.0 \times 10^{-6})|}{(0.50)^2}$$

$$F = 0.216 \, \text{N}$$

Since both charges are positive, the force is repulsive.

The Electric Field

An electric field $\vec{E}$ is a region of space where an electric charge experiences a force. It is a vector quantity.

Formula:

$$\vec{E} = \frac{\vec{F}}{q}$$

Where:

• $\vec{E}$ is the electric field strength.
• $\vec{F}$ is the electrostatic force on the charge.
• $q$ is the test charge.

Intermediate Problem

A charge $q_1 = +4.0 \times 10^{-6} \, \text{C}$ is at the origin. A second charge $q_2 = -5.0 \times 10^{-6} \, \text{C}$ is at $x = 2.0 \, \text{m}$. Find the position on the x-axis where a third charge would experience a net force of zero.

Solution

The point must be outside the charges, on the side of the smaller charge ($q_1$). Let the position be $x$.

$$k \frac{|q_1| q_3}{x^2} = k \frac{|q_2| q_3}{(2.0-x)^2}$$

$$\sqrt{\frac{4.0}{x^2}} = \sqrt{\frac{5.0}{(2.0-x)^2}}$$

$$\frac{2}{x} = \frac{\sqrt{5}}{2.0-x}$$

Solving for $x$, we get $x \approx 0.94 \, \text{m}$.

Activity: "Guess the Field"

Students, in pairs, receive cards with a charge configuration (e.g., two positive charges, a positive and negative, etc.). They must sketch the electric field lines on a mini whiteboard. Share your sketches with the class and discuss.

Simple Problem

An electron is placed in a uniform electric field $\vec{E}$ of $2.5 \times 10^4 \, \text{N/C}$. Calculate the magnitude of the force $\vec{F}$ on the electron. (Charge of electron, $e = -1.60 \times 10^{-19} \, \text{C}$)

Solution

Using the formula $F = |q|E$:

$$F = |(-1.60 \times 10^{-19}) (2.5 \times 10^4)|$$

$$F = 4.0 \times 10^{-15} \, \text{N}$$

Electric Potential

Electric potential ($V$) is the electric potential energy per unit charge at a point in an electric field. The potential difference ($\Delta V$) is the work done per unit charge to move a charge between two points.

Formula:

$$\Delta V = \frac{W}{q}$$

Intermediate Problem

A proton is released from rest in a uniform electric field $\vec{E}$ of $3.0 \times 10^5 \, \text{N/C}$. What is the change in the proton's kinetic energy after it travels $0.10 \, \text{m}$? (Charge of proton, $e = +1.60 \times 10^{-19} \, \text{C}$)

Solution

The force on the proton is $\vec{F} = q\vec{E}$. The work done is $W = Fd$. The work-energy theorem states that $W = \Delta K$.

$$\Delta K = W = Fd = (qE)d$$

$$\Delta K = (1.60 \times 10^{-19}) (3.0 \times 10^5) (0.10)$$

$$\Delta K = 4.8 \times 10^{-15} \, \text{J}$$

Activity: Peer Teaching

Form small groups. Each group is assigned a topic: (1) Conservation of Charge, (2) Charging by Induction, (3) Coulomb's Law, (4) Electric Field Lines. They will then prepare a short, 3-minute explanation to teach the rest of the class.

Summary

• Electric charge is a fundamental property of matter.
• Charges are quantized and conserved.
• Conductors allow charge to flow; insulators do not.
• Objects can be charged by friction, conduction, or induction.
• Coulomb's Law describes the force between charges.
• Electric fields are regions of force around charges.