Three point charges are located at the corners of an equilateral triangle as in the figure below. Find the magnitude and direction of the net electric force on the 1.65 μC charge. (Let A = 1.65 μC, B = 6.60 μC, and
C = −4.20 μC.)
To find the net electric force on charge A, we will first calculate the electric force between charges A and B, and the electric force between charges A and C, separately. The net electric force on charge A will be the vector sum of these two forces.
We can use the formula for electric force between two point charges: F = k|q₁q₂|/r², where k is the electrostatic constant, q₁ and q₂ are the two charges, and r is the distance between them.
Given, k = 8.99 × 10^9 N·m²/C².
First, let's calculate the electric force between charges A and B.
Charge A (q₁): 1.65 μC = 1.65 × 10^(-6) C
Charge B (q₂): 6.60 μC = 6.60 × 10^(-6) C
Distance (r): Given that they are located at the vertices of an equilateral triangle, let's assume the side length of the triangle to be s.
F_AB = k|q₁q₂|/r²
F_AB = 8.99 × 10^9 × |(1.65 × 10^(-6))(6.60 × 10^(-6))|/(s²)
Let's keep the value of F_AB in terms of 's' for now, and move on to calculate the electric force between charges A and C.
Charge C (q₂): -4.20 μC = -4.20 × 10^(-6) C
F_AC = k|q₁q₂|/r²
F_AC = 8.99 × 10^9 × |(1.65 × 10^(-6))(-4.20 × 10^(-6))|/(s²)
Now we have the magnitudes of the forces F_AB and F_AC acting on charge A. To find the net force, we need to find the sum of these two forces as vectors. Since charges A and B as well as A and C are located at the vertices of an equilateral triangle, we know that the angle between the forces F_AB and F_AC is 120°.
Let F_AN be the net force:
F_AN^2 = F_AB^2 + F_AC^2 + 2F_AB × F_AC × cos(120°)
Plugging in our values for F_AB and F_AC, we get:
F_AN^2 = (8.99 × 10^9 × |(1.65 × 10^(-6))(6.60 × 10^(-6))|/(s²))^2 + (8.99 × 10^9 × |(1.65 × 10^(-6))(-4.20 × 10^(-6))|/(s²))^2 - (8.99 × 10^9 × |(1.65 × 10^(-6))(6.60 × 10^(-6))|/(s²)) × (8.99 × 10^9 × |(1.65 × 10^(-6))(-4.20 × 10^(-6))|/(s²))
Solving for F_AN^2, we get:
F_AN = √(s² × (1.089 × 10^(-3))^2 + s² × (1.226 × 10^(-3))^2 - 2 × s² × 1.089 × 10^(-3) × 1.226 × 10^(-3))
F_AN = √(5.970 × 10^(-5) + 8.074 × 10^(-5) - 4.799 × 10^(-5))
F_AN = √(9.245 × 10^(-5) s²)
Now, we cannot solve for F_AN without the value of 's'. However, if we are given the side length of the triangle or the distance between the charges, we can find the magnitude of the net electric force.
As for the direction, we can find the angle of F_AN with respect to F_AB since we know the magnitudes and angle between F_AB and F_AC.
Let θ be the angle between F_AB and F_AN.
Using the law of cosines,
cos(θ) = [(1.089 × 10^(-3))^2 + (9.245 × 10^(-5))^2 - (1.226 × 10^(-3))^2]/(2 × 1.089 × 10^(-3) × 9.245 × 10^(-5))
cos(θ) ≈ 0.455
θ ≈ arccos(0.455) ≈ 63°
So, the direction of the net electric force F_AN is approximately 63° with respect to the direction of F_AB.
To find the magnitude and direction of the net electric force on the 1.65 μC charge, we need to determine the individual electric forces exerted on it by each of the other charges, and then vectorially add those forces.
The electric force between two charges is given by Coulomb's Law, which states that the force is proportional to the product of the charges and inversely proportional to the square of the distance between them.
Let's label the charges:
Charge A (1.65 μC) is located at the top vertex of the equilateral triangle.
Charge B (6.60 μC) is located at the bottom left vertex.
Charge C (-4.20 μC) is located at the bottom right vertex.
To calculate the magnitude of the net electric force, we need to calculate the individual forces exerted on charge A by charges B and C.
Step 1: Calculate the force between charge A and charge B:
The distance between them is the length of one side of the equilateral triangle, which we'll call d.
The formula for the electric force between two charges is:
F = k * |Q1 * Q2| / r^2
Where:
F is the magnitude of the force
k is the electrostatic constant (approximately 9 x 10^9 N m^2/C^2)
|Q1| and |Q2| are the magnitudes of the charges
r is the distance between the charges
Plugging in the values:
F_AB = (9 x 10^9 N m^2/C^2) * |1.65 μC * 6.60 μC| / d^2
Step 2: Calculate the force between charge A and charge C:
Using the same formula, we have:
F_AC = (9 x 10^9 N m^2/C^2) * |1.65 μC * (-4.20 μC)| / d^2
Step 3: Vectorially add the forces:
To find the net force on charge A, we need to add the forces F_AB and F_AC. Since the charges are arranged in an equilateral triangle, the forces will have equal magnitudes but different directions.
The magnitude of the net force is given by the Pythagorean theorem:
|F_net| = sqrt(F_AB^2 + F_AC^2)
The direction of the net force can be found using trigonometry. Since the forces make angles of 120 degrees with each other (due to the equilateral triangle), we can use the law of cosines to determine the direction angle.
θ = arccos[(F_AC^2 + F_AB^2 - F_net^2) / (2 * F_AB * F_AC)]
Once you have the magnitude and direction of the net electric force, you will have the answer to the problem.
To find the magnitude and direction of the net electric force on the 1.65 μC charge, we can use Coulomb's Law, which states that the force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Let's label the charges: A = 1.65 μC (the charge we are interested in), B = 6.60 μC, and C = -4.20 μC.
Step 1: Find the individual forces between charge A and the other charges.
The electric force between two charges is given by:
F = (k * |q1 * q2|) / r^2
Where:
F is the force between the charges.
k is the electrostatic constant, approximately 9 x 10^9 Nm^2/C^2.
q1 and q2 are the magnitudes of the charges.
r is the distance between the charges.
Let's calculate the force between charge A and charge B:
F_AB = (k * |qA * qB|) / r_AB^2
The distances between charges in an equilateral triangle are equal. Let's denote this distance as 'r'.
The distance between charge A and B is the side length of the equilateral triangle. Let's denote it as 'L'.
Since it's an equilateral triangle, we know that the distance between any two charges is equal to L. Therefore, r_AB = L.
Let's substitute the values into the formula:
F_AB = (9 x 10^9 Nm^2/C^2) * |(1.65 x 10^-6 C) * (6.60 x 10^-6 C)| / L^2
Similarly, let's calculate the force between charge A and charge C:
F_AC = (9 x 10^9 Nm^2/C^2) * |(1.65 x 10^-6 C) * (-4.20 x 10^-6 C)| / L^2
Step 2: Find the net electric force on charge A.
The net force is the vector sum of the individual forces. Since forces are vectors, we need to consider both magnitude and direction.
Let's calculate the net force:
Net Force (F_net) = F_AB + F_AC
To find the direction, we need to consider the angles between the individual forces and the x-axis. Since the charges are located at the corners of an equilateral triangle, the angle between any two charges is 120 degrees.
Assuming the x-axis is horizontal, we can determine the components of the forces. Let's denote the angle between the x-axis and F_AB as θ_AB, and the angle between the x-axis and F_AC as θ_AC.
The x-component X_AB of the force F_AB is given by:
X_AB = F_AB * cos(θ_AB)
The y-component Y_AB of the force F_AB is given by:
Y_AB = F_AB * sin(θ_AB)
Similarly, we can find the x-component X_AC and the y-component Y_AC of the force F_AC.
The net x-component of the force will be the sum of the x-components of F_AB and F_AC, and the net y-component will be the sum of the y-components of F_AB and F_AC.
Finally, the magnitude of the net force can be found using the Pythagorean theorem as the square root of the sum of the squares of the x and y components.
The direction of the net force can be found by taking the inverse tangent (tan^(-1)) of the y-component over the x-component.
By following these steps, you can calculate the magnitude and direction of the net electric force on the 1.65 μC charge using Coulomb's Law.