Three point charges lie on the x-axis as shown. What is the magnitude and direction of the electric field created at the position x=4.00 m by these charges?
q1= +4.00 C
q2=-3.00 C
q3= +2.00 C
To find the magnitude and direction of the electric field created at the position x=4.00 m, we need to calculate the electric field contribution from each point charge and then sum them up.
Let's define the positive direction of the x-axis to be towards the right.
The electric field created by a point charge is given by the equation:
E = k * q / r^2
where:
E is the magnitude of the electric field,
k is Coulomb's constant (k = 9.0 x 10^9 Nm^2/C^2),
q is the charge of the point charge,
r is the distance from the point charge to the point where we want to measure the electric field.
First, let's consider q1 = +4.00 C, which is located at x1 = 1.00 m.
The distance from q1 to the point x = 4.00 m is: r1 = 4.00 m - 1.00 m = 3.00 m.
The electric field created by q1 at x = 4.00 m is:
E1 = k * q1 / r1^2 = (9.0 x 10^9 Nm^2/C^2) * (4.00 C) / (3.00 m)^2
Next, let's consider q2 = -3.00 C, which is located at x2 = -2.00 m.
The distance from q2 to the point x = 4.00 m is: r2 = 4.00 m - (-2.00 m) = 6.00 m.
The electric field created by q2 at x = 4.00 m is:
E2 = k * q2 / r2^2 = (9.0 x 10^9 Nm^2/C^2) * (-3.00 C) / (6.00 m)^2
Finally, let's consider q3 = +2.00 C, which is located at x3 = 6.00 m.
The distance from q3 to the point x = 4.00 m is: r3 = 4.00 m - 6.00 m = -2.00 m.
The electric field created by q3 at x = 4.00 m is:
E3 = k * q3 / r3^2 = (9.0 x 10^9 Nm^2/C^2) * (2.00 C) / (-2.00 m)^2
Now, we can sum up the electric fields created by each charge:
E_total = E1 + E2 + E3
Calculate the magnitudes and directions of E1, E2, and E3 using the given values of q1, q2, q3, and the equations above.
To calculate the magnitude and direction of the electric field created at a specific position, we can use the principle of superposition, which states that the total electric field at a point is the vector sum of the individual electric fields created by each charge.
Let's calculate the electric fields created by each charge at position x=4.00 m and then add them up:
1. For charge q1= +4.00 C:
The electric field created by a point charge can be calculated using the formula:
Electric field (E1) = (k * q1) / r1^2,
where k is the electrostatic constant (k = 9.0 x 10^9 Nm^2/C^2) and r1 is the distance from q1 to the point of interest.
In this case, the distance from q1 to x=4.00 m is r1 = 4.00 m.
Plugging in the values, we find:
E1 = (9.0 x 10^9 Nm^2/C^2 * 4.00 C) / (4.00 m)^2
2. For charge q2 = -3.00 C:
Similarly, we can calculate the electric field due to q2 at x=4.00 m.
The distance from q2 to x=4.00 m is r2 = 4.00 m.
E2 = (9.0 x 10^9 Nm^2/C^2 * -3.00 C) / (4.00 m)^2
3. For charge q3 = +2.00 C:
The distance from q3 to x=4.00 m is r3 = 4.00 m.
E3 = (9.0 x 10^9 Nm^2/C^2 * 2.00 C) / (4.00 m)^2
Now, we can sum up the electric fields created by each charge to find the total electric field at x=4.00 m:
Total electric field (Et) = E1 + E2 + E3
To determine the direction of the electric field, we need to consider the sign of the charges and the direction of the distances. The electric field points in the direction that a positive test charge would experience a force.
If the resulting electric field value is positive, the field points away from the charges and if the resulting field value is negative, the field points towards the charges.
To calculate the magnitude and direction of the electric field created at the position x=4.00 m by these charges, we need to use the principle of superposition. The principle of superposition states that the total electric field at a point due to multiple charges is the vector sum of the individual electric fields at that point created by each charge.
First, let's calculate the electric field created by each charge at the point x=4.00 m. The electric field at a point due to a point charge is given by the formula:
E = (k * q) / r^2
Where:
- E is the electric field
- k is the Coulomb's constant (k = 8.99 x 10^9 N m^2/C^2)
- q is the charge
- r is the distance between the charge and the point where the electric field is being calculated
Electric field created by q1:
q1 = +4.00 C
r1 = (4.00 - 0) m = 4.00 m
E1 = (k * q1) / r1^2
Electric field created by q2:
q2 = -3.00 C
r2 = (4.00 - (-1.00)) m = 5.00 m
E2 = (k * q2) / r2^2
Electric field created by q3:
q3 = +2.00 C
r3 = (4.00 - 6.00) m = 2.00 m
E3 = (k * q3) / r3^2
Next, we need to find the vector sum of these electric fields using the principle of superposition. Since q1 and q3 are positive charges and q2 is a negative charge, the electric fields created by q1 and q3 will point away from these charges, while the electric field created by q2 will point towards it.
Add the vector components of each electric field separately. The x-component of each electric field can be calculated using Coulomb's Law:
Ex = E * cos(theta)
Where:
- Ex is the x-component of the electric field
- E is the magnitude of the electric field
- theta is the angle between the electric field direction and the x-axis
To calculate the magnitude and direction of the total electric field at x=4.00 m, we need to sum up the x-components of the individual electric fields. The y-components will cancel out due to symmetry, as the charges lie on the x-axis.
Ex_total = (Ex1 + Ex2 + Ex3)
Finally, calculate the magnitude and direction of the total electric field:
Magnitude: |E_total| = sqrt(Ex_total^2)
Direction: If Ex_total is positive, the total electric field points in the positive x-direction. If Ex_total is negative, the total electric field points in the negative x-direction.
Plugging in the values for q1, q2, q3, r1, r2, and r3 into the equations and performing the calculations will give you the final answer.