Math Problem Statement

using the image, find the x-component of the electric field at the origin, point O.

Using the image, find the x-component of the electric field at the origin, point O.

Solution

To find the x-component of the electric field at the origin (O) due to the two point charges, we need to apply Coulomb's law and consider the contributions from both charges.

Given Data:

  • q1=8.00nC=8.00×109Cq_1 = 8.00 \, \text{nC} = 8.00 \times 10^{-9} \, \text{C} at x=+16.0mx = +16.0 \, \text{m}
  • q2=6.00nC=6.00×109Cq_2 = 6.00 \, \text{nC} = 6.00 \times 10^{-9} \, \text{C} at x=9.00mx = -9.00 \, \text{m}

Coulomb's Law gives the magnitude of the electric field due to a point charge: E=keqr2E = k_e \frac{|q|}{r^2} where:

  • EE is the electric field,
  • ke=8.99×109Nm2/C2k_e = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2 / \text{C}^2 is the Coulomb constant,
  • qq is the charge,
  • rr is the distance from the charge to the point of interest.

Step 1: Electric Field Due to q1q_1

The distance from q1q_1 to the origin is r1=16.0mr_1 = 16.0 \, \text{m}. Using Coulomb's law: E1=keq1r12=8.99×1098.00×109(16.0)2E_1 = k_e \frac{|q_1|}{r_1^2} = 8.99 \times 10^9 \cdot \frac{8.00 \times 10^{-9}}{(16.0)^2} E1=2.81N/CE_1 = 2.81 \, \text{N/C} The electric field due to q1q_1 points toward the negative x-direction because the field direction is away from the positive charge at +16.0m+16.0 \, \text{m}.

Step 2: Electric Field Due to q2q_2

The distance from q2q_2 to the origin is r2=9.00mr_2 = 9.00 \, \text{m}. Using Coulomb's law: E2=keq2r22=8.99×1096.00×109(9.00)2E_2 = k_e \frac{|q_2|}{r_2^2} = 8.99 \times 10^9 \cdot \frac{6.00 \times 10^{-9}}{(9.00)^2} E2=6.66N/CE_2 = 6.66 \, \text{N/C} The electric field due to q2q_2 points toward the positive x-direction because the field direction is toward the negative charge at 9.00m-9.00 \, \text{m}.

Step 3: Combine the Electric Fields

Now we sum the x-components of the electric fields. Both fields are along the x-axis, but they point in opposite directions:

  • E1E_1 points in the negative x-direction (so it contributes 2.81N/C-2.81 \, \text{N/C}),
  • E2E_2 points in the positive x-direction (so it contributes +6.66N/C+6.66 \, \text{N/C}).

Thus, the total x-component of the electric field at the origin is: Ex=E2E1=6.66N/C2.81N/C=3.85N/CE_x = E_2 - E_1 = 6.66 \, \text{N/C} - 2.81 \, \text{N/C} = 3.85 \, \text{N/C}

Final Answer:

The x-component of the electric field at the origin is 3.85 N/C in the positive x-direction.

Would you like any additional details or have further questions?

Here are 5 related questions to deepen your understanding:

  1. How would the result change if both charges were negative?
  2. What would the y-component of the electric field be at the origin?
  3. How does the distance between the charges affect the electric field strength?
  4. What would happen if a third charge were placed at the origin?
  5. How would you calculate the force on a third charge placed at the origin?

Tip: Always keep track of the direction of electric field vectors, as their direction is crucial in determining the net field.

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Math Problem Analysis

Mathematical Concepts

Electric Fields
Coulomb's Law
Vector Addition

Formulas

E = k_e * |q| / r^2
E_x = E_2 - E_1

Theorems

Coulomb's Law

Suitable Grade Level

Undergraduate Physics