Physics M.5: Electricity Problems With Solutions

Hey guys! Physics can be a tricky subject, especially when you dive into the world of electricity. But don't worry, we're here to help you tackle those challenging problems. In this article, we'll explore some common electricity problems that Mathayom 5 students often encounter, complete with detailed solutions. Get ready to boost your understanding and ace your next physics exam!

Understanding Electric Charge and Coulomb's Law

Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electromagnetic field. The concept of electric charge is central to understanding electricity. These charges come in two forms: positive (protons) and negative (electrons). Objects with the same type of charge repel each other, while objects with opposite charges attract. The unit of electric charge is the coulomb (C). Understanding these fundamental concepts is the foundation upon which all electrical knowledge is built.

Coulomb's Law, a cornerstone of electrostatics, quantifies the force between two point charges. This law states that the force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:

F = k * (|q1 * q2|) / r^2

Where:

• F is the electrostatic force
• k is Coulomb's constant (approximately 8.99 x 10^9 N⋅m^2/C2)
• q1 and q2 are the magnitudes of the charges
• r is the distance between the charges

Let's tackle a problem to illustrate this. Suppose we have two charges, one of +2 microcoulombs and another of -3 microcoulombs, separated by a distance of 10 cm. What is the force between them?

First, convert the charges to coulombs: q1 = 2 x 10^-6 C and q2 = -3 x 10^-6 C. Convert the distance to meters: r = 0.1 m. Now, plug these values into Coulomb's Law:

F = (8.99 x 10^9 N⋅m^2/C2) * (|2 x 10^-6 C * -3 x 10^-6 C|) / (0.1 m)^2 F ≈ -5.39 N

The negative sign indicates that the force is attractive, as expected between opposite charges. Understanding how to apply Coulomb's Law is crucial for solving a variety of electrostatic problems.

Understanding electric fields is also crucial. An electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects. The direction of the electric field is defined as the direction of the force on a positive test charge. Electric field lines are used to visualize electric fields, with the density of the lines indicating the strength of the field. A stronger field exerts a greater force on any charge within it.

Problem 1: Finding the Electric Force

Question: Two small spheres are 20 cm apart and have equal charges. How many electrons must be added to each sphere so that the force of repulsion between them is 4.57 x 10^-22 N?

Solution:

Let's denote the charge on each sphere as 'q'. According to Coulomb's Law:

F = k * (q^2) / r^2

Where:

• F = 4.57 x 10^-22 N
• r = 0.20 m
• k = 8.99 x 10^9 N⋅m^2/C2

Rearranging the formula to solve for q:

q = sqrt((F * r^2) / k) q = sqrt((4.57 x 10^-22 N * (0.20 m)^2) / (8.99 x 10^9 N⋅m^2/C2)) q ≈ 4.51 x 10^-17 C

Now, to find the number of electrons (n) needed to produce this charge, we use the elementary charge (e = 1.602 x 10^-19 C):

n = q / e n = (4.51 x 10^-17 C) / (1.602 x 10^-19 C) n ≈ 281.5

Since we can't have a fraction of an electron, we round it to the nearest whole number. Therefore, approximately 282 electrons must be added to each sphere.

Delving into Electric Potential and Potential Difference

Electric potential, often referred to as voltage, represents the amount of work needed to move a unit of positive charge from a reference point to a specific point in an electric field. It is a scalar quantity measured in volts (V). The electric potential at a point is dependent on the electric field and the position of the point relative to the source charges creating the field. Understanding electric potential is crucial for analyzing circuits and the behavior of charges in electric fields.

Potential difference, on the other hand, is the difference in electric potential between two points. It is the driving force behind the movement of charge in a circuit. When there is a potential difference between two points, charges will naturally flow from the point of higher potential to the point of lower potential, creating an electric current. Batteries and power supplies are common sources of potential difference in circuits. The greater the potential difference, the stronger the push on the charges and the larger the resulting current.

The relationship between electric potential (V), work (W), and charge (q) is given by:

V = W / q

This equation signifies that the electric potential is the work done per unit charge. If it takes 10 Joules of work to move a 2 Coulomb charge from point A to point B, the potential difference between A and B is 5 Volts.

Let's consider another problem. Suppose a charge of 5 microcoulombs is moved from a point where the electric potential is 20 V to a point where it is 50 V. How much work is done?

Using the formula W = q * V, where V is the potential difference (50 V - 20 V = 30 V):

W = (5 x 10^-6 C) * (30 V) W = 1.5 x 10^-4 J

Therefore, 1.5 x 10^-4 Joules of work are done in moving the charge.

Problem 2: Calculating Electric Potential

Question: How much work is required to carry a charge of 6 μC from a point with a potential of 100 V to a point with a potential of 30 V?

Solution:

The potential difference (ΔV) is:

ΔV = 30 V - 100 V = -70 V

The work (W) required is given by:

W = q * ΔV W = (6 x 10^-6 C) * (-70 V) W = -4.2 x 10^-4 J

The negative sign indicates that the work is done by the electric field, not against it.

Exploring Capacitance and Capacitors

Capacitance is the ability of a component or circuit to collect and store energy in the form of an electrical charge. Capacitance is defined as the ratio of the change in an electric charge in a system to the corresponding change in its electric potential. A capacitor is a passive electronic component that stores electrical energy in an electric field. In its simplest form, a capacitor consists of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied across the plates, an electric field forms between them, causing charge to accumulate on the plates. The amount of charge that can be stored for a given voltage is determined by the capacitance of the capacitor.

The capacitance (C) is determined by:

C = Q / V

Where:

• Q is the charge stored
• V is the voltage across the capacitor

The unit of capacitance is the farad (F). A 1-farad capacitor can store 1 coulomb of charge at a potential difference of 1 volt.

The energy (U) stored in a capacitor is given by:

U = 0.5 * C * V^2

This equation shows that the energy stored is proportional to both the capacitance and the square of the voltage.

Consider a 100 microfarad capacitor charged to a voltage of 20 V. How much energy is stored?

U = 0.5 * (100 x 10^-6 F) * (20 V)^2 U = 0.02 J

Therefore, the capacitor stores 0.02 Joules of energy.

Problem 3: Calculating Capacitance

Question: A parallel-plate capacitor has a capacitance of 6 μF when filled with air. When a dielectric is inserted between the plates, the capacitance becomes 30 μF. What is the dielectric constant (κ) of the material?

Solution:

The dielectric constant (κ) is the ratio of the capacitance with the dielectric (C') to the capacitance without the dielectric (C):

κ = C' / C κ = (30 μF) / (6 μF) κ = 5

Therefore, the dielectric constant of the material is 5.

Understanding Electric Current and Resistance

Electric current is the rate of flow of electric charge past a point or region. In simple terms, it is the movement of electrons through a conductor. The magnitude of the current is defined as the amount of charge flowing per unit of time. Current is typically measured in amperes (A), where 1 ampere is equal to 1 coulomb of charge flowing per second.

Resistance, on the other hand, is the opposition to the flow of electric current in a circuit. It is a measure of how difficult it is for current to pass through a material or component. Resistance arises from the collisions between electrons and the atoms within the conductor, which impede their movement. The higher the resistance, the lower the current for a given voltage.

Ohm's Law establishes a relationship between voltage (V), current (I), and resistance (R):

V = I * R

This law states that the voltage across a conductor is directly proportional to the current flowing through it, with the constant of proportionality being the resistance. Ohm's Law is fundamental to analyzing and designing electrical circuits.

Let's solve a problem. Suppose a resistor with a resistance of 10 ohms is connected to a 12-volt battery. What is the current flowing through the resistor?

Using Ohm's Law: I = V / R

I = 12 V / 10 Ω I = 1.2 A

Therefore, the current flowing through the resistor is 1.2 amperes.

Problem 4: Applying Ohm's Law

Question: A 30-ohm resistor is connected to a 9-volt battery. What is the current flowing through the resistor?

Solution:

Using Ohm's Law (V = IR):

I = V / R I = (9 V) / (30 Ω) I = 0.3 A

Therefore, the current flowing through the resistor is 0.3 amperes.

Series and Parallel Circuits

Series circuits and parallel circuits represent two fundamental ways of connecting components in an electrical circuit. In a series circuit, components are connected end-to-end along a single path. This means that the current flowing through each component is the same. The total resistance in a series circuit is the sum of the individual resistances:

Req = R1 + R2 + R3 + ...

The voltage across each component in a series circuit is different and depends on its resistance. The sum of the voltages across each component equals the total voltage supplied by the source.

In contrast, in a parallel circuit, components are connected across each other, providing multiple paths for the current to flow. The voltage across each component in a parallel circuit is the same, equal to the voltage supplied by the source. However, the current flowing through each component can be different, depending on its resistance. The reciprocal of the total resistance in a parallel circuit is the sum of the reciprocals of the individual resistances:

1/Req = 1/R1 + 1/R2 + 1/R3 + ...

Understanding the characteristics of series and parallel circuits is essential for designing and analyzing more complex electrical systems.

Let's consider a series circuit with two resistors, R1 = 5 ohms and R2 = 10 ohms, connected to a 15-volt battery. What is the current flowing through the circuit?

First, find the equivalent resistance: Req = 5 Ω + 10 Ω = 15 Ω

Then, use Ohm's Law: I = V / Req I = 15 V / 15 Ω I = 1 A

Therefore, the current flowing through the circuit is 1 ampere.

Problem 5: Analyzing Series Circuits

Question: Three resistors with values of 2 Ω, 4 Ω, and 6 Ω are connected in series to a 24 V battery. Find the total current in the circuit.

Solution:

First, find the equivalent resistance (Req) of the series circuit:

Req = R1 + R2 + R3 Req = 2 Ω + 4 Ω + 6 Ω Req = 12 Ω

Now, use Ohm's Law to find the total current (I):

V = I * Req I = V / Req I = (24 V) / (12 Ω) I = 2 A

Therefore, the total current in the circuit is 2 amperes.

Conclusion

Alright guys, that wraps up our deep dive into electricity problems for Mathayom 5 students! We've covered essential concepts like electric charge, Coulomb's Law, electric potential, capacitance, current, resistance, and circuit analysis. By understanding these principles and practicing with example problems, you'll be well-equipped to tackle any electricity challenge that comes your way. Keep practicing, and you'll master these concepts in no time! Good luck with your studies!