Answer:
The y component is: Ey
Step-by-step explanation:
To calculate the electric field at point P, we need to use the principle of superposition, which states that the net electric field at a point due to a group of charges is the vector sum of the electric fields at that point due to each individual charge.
The electric field at point P due to the negative charge (-Q) is given by:
E1 = k(-Q)/r1^2
where k is Coulomb's constant, r1 is the distance between the negative charge and point P. Since the negative charge is located on the negative x-axis, the distance r1 is given by:
r1 = sqrt(a^2 + d^2)
where d is the distance between the negative charge and the origin, and is equal to:
d = -x1 = -a
So we have:
r1 = sqrt(a^2 + (-a)^2) = sqrt(2a^2)
Plugging this into the equation for E1, we get:
E1 = k(-Q)/(2a^2)
The electric field at point P due to each of the positive charges (+Q) is given by:
E2 = k(+Q)/r2^2
where r2 is the distance between the positive charge and point P. Since the positive charges are located on the positive x-axis, the distances r2 are given by:
r2 = sqrt(a^2 + (d ± x2)^2)
where d is the distance between the negative charge and the origin, x2 is the distance between the positive charge and the origin, and the ± sign corresponds to the two positive charges. Since the charges are equally spaced, we have:
x2 = d/2 = -a/2
Plugging this into the equation for r2, we get:
r2 = sqrt(a^2 + (a/2 ± a/2)^2) = sqrt(5a^2/4)
Plugging this into the equation for E2, we get:
E2 = k(+Q)/(5a^2/4) = 4k(+Q)/(5a^2)
To find the net electric field at point P, we need to add up the contributions from each charge. Since the negative charge is located on the negative x-axis, its contribution to the electric field at point P is purely in the y direction. The contributions from the two positive charges are at angles θ1 and θ2 with respect to the positive y-axis, respectively. These angles can be found using trigonometry:
tan(θ1) = a/2d = -1/2
tan(θ2) = -a/2d = 1/2
So we have:
θ1 = -26.6°
θ2 = 26.6°
The x and y components of the electric field due to each charge can be found using:
Ex = E cosθ
Ey = E sinθ
where E is the magnitude of the electric field due to the charge. The x and y components of the net electric field are then found by adding up the x and y components due to each charge.
The x component of the electric field due to the negative charge is zero, since it points along the negative x-axis. The y component is given by:
Ey1 = E1 sin(90°) = E1
The x component of the electric field due to each positive charge is given by:
Ex2 = E2 cosθ2
The y component is given by:
Ey2 = E2 sinθ2
So the x component of the net electric field at point P is:
Ex = Ex2 + Ex2 = 2E2 cosθ2
The y component is:
Ey