Two identical negative charges are located on the y axis, at y = +4m and y = -4m. What is the direction of the net electric field at a point on the x axis at x = 2m?
To find the direction of the net electric field at a point on the x axis due to two identical negative charges on the y axis, we can use the principle of superposition.
Step 1: Identify the charges and their positions
We have two identical negative charges located on the y axis, one at y = +4m and the other at y = -4m.
Step 2: Determine the electric field due to each charge
The electric field due to a point charge can be calculated using the formula:
E = k(q / r^2)
Where:
E is the electric field
k is the electrostatic constant (= 9 x 10^9 Nm^2/C^2)
q is the magnitude of the charge
r is the distance from the charge to the point of interest.
In this case, since the charges are identical, they have the same magnitude, so we can write q for both of them.
Step 3: Calculate the electric field due to each charge at the point on the x axis
The electric field due to the first charge can be calculated as:
E1 = k(q / r1^2)
Where r1 is the distance from the first charge (y = +4m) to the point on the x axis (x = 2m).
The electric field due to the second charge can be calculated as:
E2 = k(q / r2^2)
Where r2 is the distance from the second charge (y = -4m) to the point on the x axis (x = 2m).
Step 4: Determine the direction of the net electric field
Since the two charges are negative, the electric field vectors due to each charge will point towards the charges.
At the point on the x axis, the positive y component of the electric field due to the top charge (E1) and the negative y component of the electric field due to the bottom charge (E2) will cancel each other out.
Therefore, the net electric field at the point on the x axis will only have an x component, which is pointing away from the charges.
To find the direction of the net electric field at a point on the x-axis, you need to analyze the contributions of the two charges located on the y-axis.
The net electric field is the vector sum of the electric fields created by each charge at the point of interest.
Let's consider the charges. Since they are identical negative charges, they will both create electric fields directed away from themselves.
At a point on the x-axis at x = 2m, you need to evaluate the electric fields created by each charge and determine their direction.
The electric field created by a point charge is given by the formula:
E = k * (Q / r^2) * u
where E is the electric field, k is the electrostatic constant, Q is the charge magnitude, r is the distance between the charge and the point of interest, and u is the unit vector pointing from the charge to the point of interest.
Let's denote the positive charge at y = +4m as Q1 and the negative charge at y = -4m as Q2. They have the same charge magnitude Q.
The distance between the charges and the point on the x-axis can be found using the Pythagorean theorem:
d = sqrt(x^2 + y^2)
For the point on the x-axis at x = 2m, we have:
d1 = sqrt(2^2 + 4^2) = sqrt(4 + 16) = sqrt(20) m
d2 = sqrt(2^2 + (-4)^2) = sqrt(4 + 16) = sqrt(20) m
Now, we can calculate the electric fields created by each charge at the point on the x-axis:
E1 = k * (Q / d1^2) * u1
E2 = k * (Q / d2^2) * u2
The unit vectors u1 and u2 will have opposite directions since the charges are located on the opposite sides of the x-axis.
After calculating the respective electric fields, you can find the net electric field by summing up the two electric fields vectorially:
E_net = E1 + E2
The direction of the net electric field will be the direction of the resultant vector, which you can determine by comparing the magnitudes and directions of E1 and E2.
So, analyze the magnitudes and directions of the electric fields created by each charge, and then sum them up to find the direction of the net electric field at that point on the x-axis.