A 17.3-nC point charge is placed on the x-axis at x = 4.05 m, and a −35.5-nC point charge is placed on the y-axis at y = −3.57 m. What is the direction of the electric field at the origin? (Give your answer in degrees relative to the positive x-axis.)
Question
A 17.3-nC point charge is placed on the x-axis at x = 4.05 m, and a −35.5-nC point charge is placed on the y-axis at y = −3.57 m. What is the direction of the electric field at the origin? (Give your answer in degrees relative to the positive x-axis.)
Solution
The electric field E created by a point charge q at a distance r is given by the formula:
E = k*q/r^2
where k is Coulomb's constant (8.9910^9 Nm^2/C^2).
First, we calculate the electric field at the origin due to each charge separately.
For the 17.3 nC charge at x = 4.05 m:
E1 = kq1/r1^2 = (8.9910^9 Nm^2/C^2)(17.3*10^-9 C)/(4.05 m)^2 = 9.33 N/C
This field points in the positive x direction because the charge is positive.
For the -35.5 nC charge at y = -3.57 m:
E2 = kq2/r2^2 = (8.9910^9 Nm^2/C^2)(-35.5*10^-9 C)/(-3.57 m)^2 = 24.9 N/C
This field points in the positive y direction because the charge is negative and it is located in the negative y direction.
The total electric field at the origin is the vector sum of these two fields. Since they are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the total field:
E = sqrt(E1^2 + E2^2) = sqrt((9.33 N/C)^2 + (24.9 N/C)^2) = 26.3 N/C
The direction of the total field is given by the angle θ it makes with the positive x-axis. We can find this angle using the tangent function:
tan(θ) = E2/E1 = 24.9 N/C / 9.33 N/C
θ = arctan(24.9/9.33) = 69.6 degrees
So, the direction of the electric field at the origin is 69.6 degrees relative to the positive x-axis.
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