Unit 1: Fluids: Pressure and Forces
Unit 2: Thermodynamics
Unit 3: Electric Force, Field, and Potential
Unit 4: Electric Circuits
Unit 5: Magnetism and Electromagnetic Induction
Unit 6: Geometric and Physical Optics
Unit 7: Quantum, Atomic, and Nuclear Physics
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3. Electric Systems And Charge

# Electric Systems and Charge

Understanding electric systems and charge is fundamental for mastering the principles of electricity and magnetism in the AP Physics exam. This topic covers the behavior of electric charges, electric fields, and the laws governing electric forces and potential. Below are detailed notes along with five examples to help you achieve a high score on your AP Physics exam.

## Learning Objectives

By studying Electric Systems and Charge for the AP Physics exam, you should understand the principles of electric charge, electric force, and electric fields. You should be able to analyze electric potential and potential energy in various configurations, apply Coulomb’s law, and understand the behavior of conductors and insulators. Additionally, you should grasp the concepts of electric circuits, including Ohm’s law, resistivity, and the functionality of series and parallel circuits, enabling you to solve complex circuit problems effectively.

## Electric Charge

Electric Charge (q): Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric and magnetic field. There are two types of charges: positive and negative.

Key Points:

• Like charges repel each other.
• Opposite charges attract each other.
• The unit of charge is the Coulomb (C).

### Quantization of Charge

Quantization of Charge: Charge is quantized, meaning it occurs in discrete amounts. The smallest unit of charge is the elementary charge eee, where: e = 1.602 × 10⁻¹⁹

### Conservation of Charge

Conservation of Charge: The total charge in an isolated system remains constant. Charge cannot be created or destroyed, only transferred from one object to another.

## Coulomb’s Law

Coulomb’s Law: Coulomb’s Law describes the force between two point charges. The electric force F between two charges q₁​ and q₂ separated by a distance r is given by:

$F = k_e \frac{|q_1 q_2|}{r^2}$

where:

• F is the magnitude of the force,
• ke is Coulomb’s constant (8.9875×10⁹ N⋅m²⋅C⁻²),
• q₁​ and q₂ are the charges,
• r is the distance between the charges.

Key Points:

• The force is attractive if the charges are of opposite signs and repulsive if they are of the same sign.
• The force acts along the line joining the centers of the two charges.

## Electric Field

Electric Field (E): An electric field is a region around a charged object where another charged object experiences an electric force. The electric field E due to a point charge q at a distance r is given by:

$E = k_e \frac{|q|}{r^2}$

Key Points:

• The direction of the electric field is away from positive charges and towards negative charges.
• The unit of electric field is Newtons per Coulomb (N/C).

## Electric Potential

Electric Potential (V): Electric potential is the electric potential energy per unit charge. The electric potential V at a distance r from a point charge q is given by:

$V = k_e \frac{q}{r}$

Key Points:

• The unit of electric potential is Volts (V).
• The electric potential due to multiple charges is the algebraic sum of the potentials due to individual charges.

## Examples of Electric Systems and Charge

Example 1: Calculating Electric Force

Scenario: Two charges, q₁​ = +3μC and q₂ ​= −5μC, are placed 0.2 meters apart. Calculate the electric force between them.

Solution: $F = k_e \frac{|q_1 q_2|}{r^2}$ $F = 8.9875 \times 10^9 \frac{|3 \times 10^{-6} \times -5 \times 10^{-6}|}{(0.2)^2}$ $F = 8.9875 \times 10^9 \frac{15 \times 10^{-12}}{0.04}$ $F = 8.9875 \times 10^9 \times 3.75 \times 10^{-10}$ $F = 3.37 \, \text{N}$

Example 2: Electric Field Due to a Point Charge

Scenario: Calculate the electric field at a point 0.1 meters away from a +2μC charge.

Solution: $E = k_e \frac{|q|}{r^2}$ $E = 8.9875 \times 10^9 \frac{2 \times 10^{-6}}{(0.1)^2}$ $E = 8.9875 \times 10^9 \frac{2 \times 10^{-6}}{0.01}$ $E = 8.9875 \times 10^9 \times 2 \times 10^{-4}$ $E = 1.8 \times 10^6 \, \text{N/C}$

Example 3: Electric Potential Due to a Point Charge

Scenario: Find the electric potential at a distance of 0.5 meters from a +1C charge.

Solution: $V = k_e \frac{q}{r}$ $V = 8.9875 \times 10^9 \frac{1}{0.5}$ $V = 8.9875 \times 10^9 \times 2$ $V = 1.8 \times 10^{10} \, \text{V}$

Example 4: Superposition of Electric Fields

Scenario: Two charges, q₁​ = +4μC and q₂ ​= +6μC, are placed 0.3 meters apart. Find the electric field at the midpoint between the charges.

Solution: The electric field due to q​₁ at the midpoint is:
$E_1 = k_e \frac{|q|}{(0.15)^2}$ $E_1 = 8.9875 \times 10^9 \frac{4 \times 10^{-6}}{0.0225}$ $E_1 = 1.6 \times 10^6 \, \text{N/C}$
The electric field due to q₂​ at the midpoint is:
$E_2 = k_e \frac{|q_2|}{(0.15)^2}$ $E_2 = 8.9875 \times 10^9 \frac{6 \times 10^{-6}}{0.0225}$ $E_2 = 2.4 \times 10^6 \, \text{N/C}$

Since both charges are positive, the fields are in opposite directions.
Therefore, the net field at the midpoint is:

Example 5: Potential Energy of a System of Charges

Scenario: Calculate the potential energy of a system with two charges, q₁ = +3μC and q₂​ = −3μC, separated by 0.1 meters.

Solution: $U = k_e \frac{q_1 q_2}{r}$ $U = 8.9875 \times 10^9 \frac{3 \times 10^{-6} \times -3 \times 10^{-6}}{0.1}$ $U = 8.9875 \times 10^9 \frac{-9 \times 10^{-12}}{0.1}$ $U = -8.1 \times 10^{-1}$ $U = -0.81 \, \text{J}$

## Practice Problems

Question 1:

Two point charges, q₁ ​= +3μC and q₂​ = −3μC, are placed 0.5 meters apart in a vacuum. What is the magnitude of the electric force between them? (Use Coulomb’s constant k = 8.99×109N⋅m²/C²)

A) 0.27N
B) 0.54N
C) 1.08N
D) 2.16N

Explanation:

The electric force F between two point charges is given by Coulomb’s law:

$F = k \frac{|q_1 q_2|}{r^2}$

Given:

$q_1 = +3 \, \mu \text{C} = 3 \times 10^{-6} \, \text{C}$ $q_2 = -3 \, \mu \text{C} = -3 \times 10^{-6} \, \text{C}$ $r = 0.5 \, \text{m}$ $k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$

Calculate the magnitude of the force:

$F = 8.99 \times 10^9 \left( \frac{(3 \times 10^{-6} \, \text{C})(3 \times 10^{-6} \, \text{C})}{(0.5 \, \text{m})^2} \right)$ $F = 8.99 \times 10^9 \times \frac{9 \times 10^{-12}}{0.25}$ $F = 8.99 \times 10^9 \times 36 \times 10^{-12}$ $F = 323.64 \times 10^{-3} \, \text{N}$ $F = 1.08 \, \text{N}$

Thus, the correct answer is 1.08 N.

Question 2:

What is the electric potential energy of a system of two charges, q₁ ​= 4μC and q₂ = 6μC, separated by a distance of 0.3 meters in a vacuum? (Use Coulomb’s constant k=8.99×10⁹N⋅m²/C²)

A) 0.72J
B) 0.96J
C) 1.20J
D) 1.44J

Explanation:

The electric potential energy U of a system of two point charges is given by:

$U = k \frac{q_1 q_2}{r}$

Given:

$q_1 = 4 \, \mu C = 4 \times 10^{-6} \, C$ $q_2 = 6 \, \mu C = 6 \times 10^{-6} \, C$ $r = 0.3 \, m$ $k = 8.99 \times 10^{9} \, \text{N} \cdot \text{m}^2/\text{C}^2$

Calculate the electric potential energy:

$U = 8.99 \times 10^9 \left( \frac{(4 \times 10^{-6} \, \text{C})(6 \times 10^{-6} \, \text{C})}{0.3 \, \text{m}} \right)$ $U = 8.99 \times 10^9 \times \frac{24 \times 10^{-12}}{0.3}$ $U = 8.99 \times 10^9 \times 80 \times 10^{-12}$ $U = 719.2 \times 10^{-3} \, \text{J}$ $U = 0.72 \, \text{J}$

Thus, the correct answer is 0.72 J.

Question 3:

A parallel plate capacitor is connected to a 12 V battery and has a capacitance of 5 μF. What is the charge stored on the capacitor?

A) 30μC
B) 50μC
C) 60μC
D) 120μC

Explanation:

The charge Q stored on a capacitor is given by the formula:

Q = CV

where C is the capacitance and V is the voltage.

Given:

• C = 5μF = 5×10⁻⁶F
• V = 12V

Calculate the charge:

Q = 5×10⁻⁶F×12V
Q =60×10⁻⁶C
Q = 60μC

Thus, the correct answer is 60 μC.