Find the magnitude and direction of the net electrostatic force exerted on the point charge q3 in the figure below, where q = +2.5 μC and d = 28 cm. The picture is of a square. q1=+q, q2=-2q, q3=-3q, q4=-4
You will have to add the forces as vectors.
very nice question
To find the magnitude and direction of the net electrostatic force exerted on q3, we need to calculate the forces exerted by each of the other charges (q1, q2, and q4) individually and then add them vectorially.
1. Let's start by calculating the force exerted on q3 by q1.
The formula to calculate the electrostatic force is given by Coulomb's Law:
F = k * (|q1 * q3| / r^2)
where k is the electrostatic constant (k = 9 * 10^9 N m^2/C^2), |q1 * q3| is the magnitude of the product of the charges, and r is the distance between q1 and q3.
Substitute the values given:
q1 = +q (charge of q1)
q3 = -3q (charge of q3)
r = d = 28 cm = 0.28 m (distance between q1 and q3)
Plugging in the values, we have:
F1 = 9 * 10^9 N m^2/C^2 * (|(+q) * (-3q)| / (0.28 m)^2)
F1 = 9 * 10^9 N m^2/C^2 * (3q^2 / 0.0784 m^2)
F1 = 1082q^2 N
The force exerted by q1 on q3 is equal to 1082q^2 N, directed towards q1.
2. Next, let's calculate the force exerted on q3 by q2.
Using Coulomb's Law again, substituting the values:
q2 = -2q (charge of q2)
F2 = 9 * 10^9 N m^2/C^2 * (|(-2q) * (-3q)| / (0.28 m)^2)
F2 = 9 * 10^9 N m^2/C^2 * (6q^2 / 0.0784 m^2)
F2 = 2164q^2 N
The force exerted by q2 on q3 is equal to 2164q^2 N, directed towards q2.
3. Lastly, let's calculate the force exerted on q3 by q4.
Using Coulomb's Law, substituting the values:
q4 = -4q (charge of q4)
F4 = 9 * 10^9 N m^2/C^2 * (|(-4q) * (-3q)| / (0.28 m)^2)
F4 = 9 * 10^9 N m^2/C^2 * (12q^2 / 0.0784 m^2)
F4 = 4330q^2 N
The force exerted by q4 on q3 is equal to 4330q^2 N, directed towards q4.
4. Now, let's add up these forces vectorially to find the net force.
The net electrostatic force is the vector sum of the individual forces:
Fnet = √((F1 * cos(45°) + F2 *cos(180°))^2 + (F1 * sin(45°) + F4 * sin(270°))^2)
Note: We need to consider the angles between the forces and the x-axis for the vector addition.
Plugging in the calculated values:
Fnet = √((1082q^2 * cos(45°) + 2164q^2 * cos(180°))^2 + (1082q^2 * sin(45°) + 4330q^2 * sin(270°))^2)
Fnet = √((1082q^2 * 0.707 + (-2164q^2))^2 + (1082q^2 * 0.707 + 4330q^2 * (-1))^2)
Fnet = √((765.374q^2)^2 + (3656.994q^2)^2)
Fnet = √(586502.245q^4 + 13373305242q^4)
Fnet = √(13373991744q^4)
Fnet = 3656.994q^2 N
The magnitude of the net electrostatic force exerted on q3 is equal to 3656.994q^2 N.
5. The direction of the net force can be determined considering the angle between the sum of the individual forces and the x-axis.
Tan(θ) = (F1 * sin(45°) + F4 * sin(270°)) / (F1 * cos(45°) + F2 * cos(180°))
Tan(θ) = ((1082q^2 * sin(45°) + 4330q^2 * sin(270°)) / (1082q^2 * cos(45°) + 2164q^2 * cos(180°)))
Tan(θ) = (676.193q^2 / (-2055.667q^2))
θ = Tan^(-1)(-676.193 / 2055.667)
θ = -18.90°
The direction of the net electrostatic force exerted on q3 is -18.90° (clockwise from the x-axis).
Therefore, the magnitude of the net electrostatic force exerted on q3 is 3656.994q^2 N, directed at an angle of -18.90° clockwise from the x-axis.
To find the magnitude and direction of the net electrostatic force exerted on the point charge q3, we need to calculate the individual forces between q3 and each of the other charges (q1, q2, and q4), and then calculate their vector sum.
Let's start by calculating the individual forces:
1. Force between q3 and q1:
The electrostatic force between two point charges is given by Coulomb's Law:
F = (k * |q1| * |q3|) / r^2,
where F is the force, k is the electrostatic constant (9 × 10^9 N m^2/C^2), |q1| and |q3| are the magnitudes of the charges, and r is the distance between them.
In this case, q1 = +q and q3 = -3q.
The distance between them is given as "d = 28 cm," which we should convert to meters: 0.28 m.
Plugging in the values, we get:
F1 = (9 × 10^9 N m^2/C^2) * (q) * (3q) / (0.28 m)^2.
2. Force between q3 and q2:
Using the same formula, we have:
F2 = (9 × 10^9 N m^2/C^2) * (2q) * (3q) / (d)^2.
3. Force between q3 and q4:
Again, using Coulomb's Law:
F3 = (9 × 10^9 N m^2/C^2) * (4q) * (3q) / (d)^2.
Now, let's calculate the magnitudes of these individual forces.
The magnitude of a vector is given by |F| = sqrt(Fx^2 + Fy^2),
where Fx and Fy are the horizontal and vertical components of the vector.
For F1, we only have a vertical component, as indicated by the figure. So:
|F1| = |F1y| = F1 * sin(45 degrees).
For F2 and F3, we need to calculate both horizontal and vertical components.
|F2| = sqrt(F2x^2 + F2y^2) = sqrt((F2 * cos(45 degrees))^2 + (F2 * sin(45 degrees))^2).
|F3| = sqrt(F3x^2 + F3y^2) = sqrt((F3 * cos(45 degrees))^2 + (F3 * sin(45 degrees))^2).
Next, we need to calculate the direction of each force. The direction is given by the angle, θ, that the force vector makes with the positive x-axis, measured counterclockwise.
For F1, the direction is 45 degrees above the positive x-axis.
For F2 and F3, the direction is 45 degrees below the positive x-axis.
Finally, we can add all these individual vectors to find the net force on q3. The net force is the vector sum of F1, F2, and F3.
To find the magnitude and direction of the net force, we add the horizontal and vertical components of each force vector and then use trigonometry to determine the magnitude and direction of the resultant vector.