Math Problem Statement

Calculate the net electrostatic force on charge A. Given: Charge A = +3.0 μC, Charge B = -1.5 μC, Charge C = +4.0 μC. Distances: A to B = 0.20 m, A to C = 0.60 m.

Solution

We are tasked with calculating the net electrostatic force on charge AA due to charges BB and CC. Here’s how we can approach it:

Given Data:

  • Charge A=+3.0μCA = +3.0 \, \mu C
  • Charge B=1.5μCB = -1.5 \, \mu C
  • Charge C=+4.0μCC = +4.0 \, \mu C
  • Distance between AA and B=0.20mB = 0.20 \, \text{m}
  • Distance between AA and C=0.60mC = 0.60 \, \text{m}

Formula for Electrostatic Force:

The electrostatic force between two charges is given by Coulomb's Law: F=keq1q2r2F = k_e \cdot \frac{|q_1 \cdot q_2|}{r^2} where:

  • FF is the magnitude of the force,
  • ke=8.99×109Nm2/C2k_e = 8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2 is Coulomb's constant,
  • q1q_1 and q2q_2 are the magnitudes of the charges,
  • rr is the distance between the charges.

Step-by-Step Calculation:

1. Force between AA and BB:

  • The force between AA and BB will be attractive (since AA is positive and BB is negative). FAB=ke(3.0×106)(1.5×106)(0.20)2F_{AB} = k_e \cdot \frac{|(3.0 \times 10^{-6}) \cdot (-1.5 \times 10^{-6})|}{(0.20)^2} Let’s calculate FABF_{AB}.

2. Force between AA and CC:

  • The force between AA and CC will be repulsive (since both AA and CC are positive). FAC=ke(3.0×106)(4.0×106)(0.60)2F_{AC} = k_e \cdot \frac{|(3.0 \times 10^{-6}) \cdot (4.0 \times 10^{-6})|}{(0.60)^2} Let’s calculate FACF_{AC}.

3. Determine the Net Force on AA:

  • The force between AA and BB is attractive (towards BB) along the x-axis in the negative direction.
  • The force between AA and CC is repulsive (away from CC) along the x-axis in the positive direction.

Thus, the net force will be the algebraic sum: Fnet=FACFABF_{\text{net}} = F_{AC} - F_{AB} (Note: FACF_{AC} will be positive and FABF_{AB} will be negative because of direction.)

Let me calculate these values for you.The forces calculated are:

  • Force between AA and BB (attractive): FAB=1.011NF_{AB} = 1.011 \, \text{N} (acting towards BB, negative direction).
  • Force between AA and CC (repulsive): FAC=0.300NF_{AC} = 0.300 \, \text{N} (acting away from CC, positive direction).

Net Electrostatic Force on AA:

Fnet=0.712NF_{\text{net}} = -0.712 \, \text{N} This means the net force on AA is 0.712N0.712 \, \text{N} towards charge BB (negative x-direction).

Would you like more details on how these values were calculated or need further clarifications?

Here are some related questions for practice:

  1. How would the net force change if the sign of charge CC was negative?
  2. What would happen to the net force if the distance between AA and CC doubled?
  3. How would the net force on AA be affected if BB and CC had the same magnitude of charge?
  4. What is the significance of the direction of the forces in Coulomb's law?
  5. How does the medium (vacuum or other materials) between the charges affect the force?

Tip: In electrostatic problems, always consider both the magnitude and direction of forces to determine the net effect.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Electrostatics
Coulomb's Law
Vector Addition of Forces

Formulas

Coulomb's Law: F = k_e * (|q1 * q2|) / r^2
Net Force: F_net = F_AC - F_AB

Theorems

Coulomb's Law
Superposition Principle in Electrostatics

Suitable Grade Level

Grades 11-12 (Physics)