Consider the charge distribution shown below. Three particles with negative charges q1 = q2 = - 2.25 ยตC and positive charge q3 = 6.20 ยตC sit in a circle of radius 0.500 m. The angle ๐ is 50.0ยฐ.
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(a) Find the magnitude of the net electric field at the origin.
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(b) Find the angle that the net electric field at the origin makes with the +y axis.
WebAssign will check your answer for the correct number of significant figures. ยฐ
To find the magnitude of the net electric field at the origin, we can use the principle of superposition. The net electric field at a point is the vector sum of the electric fields due to each individual charge at that point.
The formula to calculate the electric field due to a point charge is given by Coulomb's law:
E = k * (q / r^2)
Where E is the magnitude of the electric field, q is the charge, r is the distance between the charge and the point, and k is the electrostatic constant (k = 8.99 x 10^9 Nm^2/C^2).
In this case, we have three charges arranged in a circle. To simplify the calculation, we can consider their x and y components separately.
First, let's find the x-component of the net electric field at the origin. The x-component of the electric field due to a charge is given by:
E_x = E * cos(๐)
Where E is the magnitude of the electric field and ๐ is the angle between the positive x-axis and the line connecting the charge to the point.
For each charge, we can calculate the x-component of the electric field using the formula above.
E_{1x} = E * cos(๐_1) = k * (q1 / r^2) * cos(๐_1)
E_{2x} = E * cos(๐_2) = k * (q2 / r^2) * cos(๐_2)
E_{3x} = E * cos(๐_3) = k * (q3 / r^2) * cos(๐_3)
Let's calculate the x-components using the given values:
k = 8.99 x 10^9 Nm^2/C^2
q1 = q2 = -2.25 ยตC = -2.25 x 10^-6 C
q3 = 6.20 ยตC = 6.20 x 10^-6 C
r = 0.500 m
๐_1 = 0ยฐ, ๐_2 = 120ยฐ (charge 2 is at 120ยฐ from charge 1), ๐_3 = 240ยฐ (charge 3 is at 240ยฐ from charge 1)
E_{1x} = (8.99 x 10^9 Nm^2/C^2) * (-2.25 x 10^-6 C) * cos(0ยฐ)
E_{2x} = (8.99 x 10^9 Nm^2/C^2) * (-2.25 x 10^-6 C) * cos(120ยฐ)
E_{3x} = (8.99 x 10^9 Nm^2/C^2) * (6.20 x 10^-6 C) * cos(240ยฐ)
Now, we can find the net x-component of the electric field at the origin by summing up the contributions from each charge:
E_{net_x} = E_{1x} + E_{2x} + E_{3x}
Next, let's find the y-component of the net electric field at the origin. The y-component of the electric field due to a charge is given by:
E_y = E * sin(๐)
For each charge, we can calculate the y-component of the electric field using the formula above.
E_{1y} = E * sin(๐_1) = k * (q1 / r^2) * sin(๐_1)
E_{2y} = E * sin(๐_2) = k * (q2 / r^2) * sin(๐_2)
E_{3y} = E * sin(๐_3) = k * (q3 / r^2) * sin(๐_3)
Again, let's calculate the y-components using the given values:
E_{1y} = (8.99 x 10^9 Nm^2/C^2) * (-2.25 x 10^-6 C) * sin(0ยฐ)
E_{2y} = (8.99 x 10^9 Nm^2/C^2) * (-2.25 x 10^-6 C) * sin(120ยฐ)
E_{3y} = (8.99 x 10^9 Nm^2/C^2) * (6.20 x 10^-6 C) * sin(240ยฐ)
Now, we can find the net y-component of the electric field at the origin by summing up the contributions from each charge:
E_{net_y} = E_{1y} + E_{2y} + E_{3y}
Finally, the magnitude of the net electric field at the origin can be calculated using the x and y components:
E_{net} = sqrt((E_{net_x})^2 + (E_{net_y})^2)
To find the angle that the net electric field at the origin makes with the +y axis, we can use the formula:
๐_{net} = arctan(E_{net_x} / E_{net_y})
Now you have the steps to find the answers to both parts (a) and (b) of the question. Substitute the given values into the formulas and perform the calculations to find the answers.