Static Electricity Physics M5: Problems & Solutions

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Hey guys! Physics can be a tough subject, especially when we dive into the world of static electricity. For students in Mathayom 5 (M.5), understanding static electricity is super important for grasping more advanced physics concepts later on. So, let's break down some common problems you might encounter and how to solve them. Think of this as your ultimate guide to conquering static electricity problems! We'll go through each problem step-by-step, making sure you understand the why behind the how. No more rote memorization – let's get to grips with the fundamentals. By the end of this, you'll be able to tackle those tricky physics questions with confidence. Ready to jump in? Let’s do this!

Understanding Static Electricity Concepts

Before we jump into the problems, let's quickly review the key concepts of static electricity. Static electricity arises from an imbalance of electric charges within or on the surface of a material. This imbalance means there's either an excess of negative charges (electrons) or positive charges (protons). Remember, it's usually the electrons that move around, not the protons, because protons are tightly bound within the nucleus of an atom.

One of the fundamental principles governing static electricity is Coulomb's Law, which describes the force between two charged objects. 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's 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 m²/C²),
  • q1 and q2 are the magnitudes of the charges, and
  • r is the distance between the charges.

Another important concept is electric field, which is the force per unit charge exerted on a test charge at a given point in space. Electric fields are vector quantities, meaning they have both magnitude and direction. The electric field due to a point charge Q at a distance r is given by:

E = k * Q / r^2

Where:

  • E is the electric field strength.

Lastly, understanding electric potential (voltage) is crucial. Electric potential is the amount of work needed to move a unit positive charge from a reference point to a specific point in an electric field. The electric potential V due to a point charge Q at a distance r is given by:

V = k * Q / r

These core concepts form the foundation for solving static electricity problems. Make sure you have a solid grasp of them before moving on to the examples!

Example Problems and Solutions

Okay, let's dive into some example problems! We'll walk through each one step-by-step, highlighting the key concepts and formulas you'll need to solve them. Get your notebooks ready – it's practice time!

Problem 1: Coulomb's Law

Problem: Two point charges, q1 = +3 μC and q2 = -4 μC, are separated by a distance of 6 cm. Calculate the magnitude and direction of the electrostatic force between them.

Solution:

  1. Identify the given values:

    • q1 = +3 x 10^-6 C
    • q2 = -4 x 10^-6 C
    • r = 0.06 m
    • k = 8.99 x 10^9 N m²/C²

  2. Apply Coulomb's Law:

    F = k * (q1 * q2) / r^2 F = (8.99 x 10^9 N m²/C²) * ((3 x 10^-6 C) * (-4 x 10^-6 C)) / (0.06 m)^2 F = -30 N

  3. Interpret the result:

    The magnitude of the force is 30 N. The negative sign indicates that the force is attractive, since the charges have opposite signs.

Answer: The electrostatic force between the two charges is 30 N, and it is an attractive force.

Problem 2: Electric Field

Problem: A point charge of +5 nC is placed at the origin. Calculate the electric field at a point 4 cm away along the x-axis.

Solution:

  1. Identify the given values:

    • Q = +5 x 10^-9 C
    • r = 0.04 m
    • k = 8.99 x 10^9 N m²/C²

  2. Apply the electric field formula:

    E = k * Q / r^2 E = (8.99 x 10^9 N m²/C²) * (5 x 10^-9 C) / (0.04 m)^2 E = 28093.75 N/C

  3. Determine the direction:

    Since the charge is positive, the electric field points radially outward from the charge. In this case, it points along the positive x-axis.

Answer: The electric field at the point 4 cm away along the x-axis is 28093.75 N/C, directed along the positive x-axis.

Problem 3: Electric Potential

Problem: What is the electric potential at a distance of 10 cm from a point charge of -8 μC?

Solution:

  1. Identify the given values:

    • Q = -8 x 10^-6 C
    • r = 0.10 m
    • k = 8.99 x 10^9 N m²/C²

  2. Apply the electric potential formula:

    V = k * Q / r V = (8.99 x 10^9 N m²/C²) * (-8 x 10^-6 C) / (0.10 m) V = -719200 V

Answer: The electric potential at a distance of 10 cm from the point charge is -719200 V.

Problem 4: Electric Force on Multiple Charges

Problem: Three charges are arranged as follows: q1 = +2 μC is at the origin (0,0), q2 = -3 μC is at (4,0) cm, and q3 = +4 μC is at (0,3) cm. Find the net force on q1.

Solution:

This problem involves vector addition of forces. We need to calculate the force exerted on q1 by q2 and q3 separately, and then add them as vectors.

  1. Calculate the force between q1 and q2:

    • q1 = +2 x 10^-6 C
    • q2 = -3 x 10^-6 C
    • r12 = 0.04 m F12 = k * (q1 * q2) / r12^2 F12 = (8.99 x 10^9 N m²/C²) * ((2 x 10^-6 C) * (-3 x 10^-6 C)) / (0.04 m)^2 F12 = -33.71 N (attractive force, directed along the x-axis towards q2)

  2. Calculate the force between q1 and q3:

    • q1 = +2 x 10^-6 C
    • q3 = +4 x 10^-6 C
    • r13 = 0.03 m F13 = k * (q1 * q3) / r13^2 F13 = (8.99 x 10^9 N m²/C²) * ((2 x 10^-6 C) * (4 x 10^-6 C)) / (0.03 m)^2 F13 = 79.91 N (repulsive force, directed along the y-axis away from q3)

  3. Add the forces as vectors:

    • F12 has components (-33.71 N, 0 N)
    • F13 has components (0 N, 79.91 N) The net force F_net is the vector sum: F_net = F12 + F13 = (-33.71 N, 79.91 N)

  4. Calculate the magnitude and direction of the net force:

    • Magnitude: |F_net| = sqrt((-33.71 N)^2 + (79.91 N)^2) = 86.76 N
    • Direction: θ = arctan(79.91 / -33.71) = -67.11 degrees (with respect to the x-axis). Since the x-component is negative and the y-component is positive, the angle is in the second quadrant. Add 180 degrees to get the correct angle: θ = 112.89 degrees.

Answer: The net force on q1 is 86.76 N at an angle of 112.89 degrees with respect to the x-axis.

Problem 5: Electric Field due to a Charged Sphere

Problem: A solid sphere of radius 5 cm carries a uniformly distributed charge of 15 nC. Calculate the electric field at a point:

(a) 2 cm from the center of the sphere.

(b) 10 cm from the center of the sphere.

Solution:

This problem requires understanding Gauss's Law. Gauss's Law states that the electric flux through any closed surface is proportional to the enclosed electric charge:

∮ E · dA = Q_enclosed / ε0

Where:

  • E is the electric field,
  • dA is the differential area vector, and
  • ε0 is the permittivity of free space (approximately 8.854 x 10^-12 C²/N m²).

(a) Inside the sphere (r < R):

  1. Apply Gauss's Law:

    Consider a Gaussian surface as a sphere of radius r = 2 cm inside the charged sphere. The charge enclosed by this Gaussian surface (Q_enclosed) is proportional to the volume of the Gaussian sphere relative to the total volume of the charged sphere. Q_enclosed = Q_total * (r^3 / R^3) Where:

    • Q_total = 15 x 10^-9 C
    • r = 0.02 m
    • R = 0.05 m Q_enclosed = (15 x 10^-9 C) * ((0.02 m)^3 / (0.05 m)^3) = 0.96 x 10^-9 C

  2. Calculate the electric field:

    E * 4πr^2 = Q_enclosed / ε0 E = Q_enclosed / (4π ε0 r^2) E = (0.96 x 10^-9 C) / (4π * 8.854 x 10^-12 C²/N m² * (0.02 m)^2) E = 10790 N/C

Answer (a): The electric field at 2 cm from the center is approximately 10790 N/C, directed radially outward.

(b) Outside the sphere (r > R):

  1. Apply Gauss's Law:

    Consider a Gaussian surface as a sphere of radius r = 10 cm outside the charged sphere. In this case, the entire charge of the sphere is enclosed by the Gaussian surface. Q_enclosed = Q_total = 15 x 10^-9 C

  2. Calculate the electric field:

    E * 4πr^2 = Q_enclosed / ε0 E = Q_enclosed / (4π ε0 r^2) E = (15 x 10^-9 C) / (4π * 8.854 x 10^-12 C²/N m² * (0.10 m)^2) E = 13485 N/C

Answer (b): The electric field at 10 cm from the center is approximately 13485 N/C, directed radially outward.

Tips for Solving Static Electricity Problems

  • Draw Diagrams: Always start by drawing a clear diagram of the problem. This helps you visualize the charges, distances, and directions of forces and electric fields.
  • Identify Given Values: List all the given values with their correct units. This ensures you don't miss any crucial information.
  • Choose the Right Formula: Select the appropriate formula based on the problem. Is it Coulomb's Law, electric field, or electric potential? Make sure you understand what each formula represents.
  • Pay Attention to Units: Ensure all values are in consistent units (e.g., meters for distance, Coulombs for charge). Convert units if necessary.
  • Consider Direction: Electric forces and fields are vectors. Pay attention to their directions. Use vector addition to find the net force or field when multiple charges are involved.
  • Check Your Work: After solving the problem, double-check your calculations and ensure that the answer makes sense in the context of the problem.

Conclusion

And there you have it! We've covered the key concepts of static electricity and worked through several example problems. Remember, practice makes perfect, so keep solving problems to strengthen your understanding. Don't be afraid to ask for help from your teachers or classmates if you get stuck. With dedication and a solid understanding of the fundamentals, you'll be able to master static electricity and excel in your physics studies. Keep up the great work, guys!