A charge q1 = 5 nC is located at the origin of xy copedinate system, and a charge of q2 = -2 nC is located on the y-axis at y =.30m. Find the components of the electric field and its magnitude and direction at a point P with coordinates (.40m, .30m).
To find the components of the electric field at point P, we can use Coulomb's law:
Electric field (E) = (k * q) / r^2
where:
- k is Coulomb's constant, approximately equal to 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
Let's find the electric field created by q1 and q2 separately:
1. Electric field created by q1 (located at the origin):
Since q1 is located at the origin, the distance between q1 and point P is:
r1 = sqrt((0.4 - 0)^2 + (0.3 - 0)^2) = sqrt(0.4^2 + 0.3^2) = sqrt(0.16 + 0.09) = sqrt(0.25) = 0.5m
Plugging in the values into the formula:
E1 = (k * q1) / r1^2
= (9 x 10^9 Nm^2/C^2 * 5 x 10^-9 C) / (0.5m)^2
= (45 x 10^0 N) / (0.25m^2)
= 180 x 10^0 N/m^2
= 180 N/m^2
The electric field created by q1 points radially outward from the origin.
2. Electric field created by q2 (located on the y-axis):
As q2 is located on the y-axis, the distance between q2 and point P is the horizontal distance between them:
r2 = 0.4m
Plugging in the values into the formula:
E2 = (k * q2) / r2^2
= (9 x 10^9 Nm^2/C^2 * -2 x 10^-9 C) / (0.4m)^2
= (-18 x 10^0 N) / (0.16m^2)
= -112.5 N/m^2
The negative sign indicates that the electric field created by q2 points radially inward towards the y-axis.
To find the net electric field at point P:
Since the electric fields E1 and E2 are in different directions, we need to find the vector sum. We'll calculate the x-component and y-component separately:
Net Electric field (Ex) = E1 + 0 (since E1 is purely in the y-direction)
= 180 N/m^2
Net Electric field (Ey) = E2 + 0 (since E2 is purely in the x-direction)
= -112.5 N/m^2
To find the magnitude and direction of the net electric field at point P:
The magnitude of the net electric field is given by:
|E| = sqrt(Ex^2 + Ey^2)
= sqrt((180 N/m^2)^2 + (-112.5 N/m^2)^2)
≈ 213.37 N/m^2
The direction of the net electric field can be found using trigonometry:
tan(theta) = Ey / Ex
theta = arctan(Ey / Ex)
theta = arctan((-112.5 N/m^2) / (180 N/m^2))
theta ≈ -32.44°
Therefore, the components of the electric field at point P are approximately Ex = 180 N/m^2, Ey = -112.5 N/m^2, and the magnitude of the electric field is approximately 213.37 N/m^2, directed at an angle of approximately -32.44° from the x-axis.
To find the components of the electric field at point P, we can use the principle of superposition. This principle states that the electric field at a point due to multiple charges is the vector sum of the electric fields created by each individual charge.
To calculate the electric field at point P, we need to consider the electric field created by each charge separately.
Step 1: Calculate the electric field due to q1 at point P:
The electric field created by a point charge is given by Coulomb's Law:
E1 = k * (q1 / r1^2)
where k is the Coulomb's constant (8.99 x 10^9 Nm^2/C^2), q1 is the charge of the source (5 nC), and r1 is the distance between the source charge and the point P.
In this case, the distance r1 can be calculated using the Pythagorean theorem:
r1 = sqrt((x1 - xP)^2 + (y1 - yP)^2)
where (x1, y1) are the coordinates of q1 (origin) and (xP, yP) are the coordinates of point P.
Substituting the given values:
r1 = sqrt((0 - 0.40m)^2 + (0 - 0.30m)^2)
r1 = sqrt(0.16m^2 + 0.09m^2)
r1 = sqrt(0.25m^2)
r1 = 0.5m
Now, plug the values into the electric field equation:
E1 = (8.99 x 10^9 Nm^2/C^2) * (5 x 10^-9 C) / (0.5m)^2
Step 2: Calculate the electric field due to q2 at point P:
Similarly, we can calculate the electric field due to q2 using Coulomb's Law:
E2 = k * (q2 / r2^2)
where q2 is the charge of the source (-2 nC) and r2 is the distance between the source charge and point P.
For q2, the distance r2 is simply the y-coordinate difference between q2 and P, because they lie on the same axis:
r2 = y2 - yP
r2 = 0.30m - 0.30m
r2 = 0
Since r2 is 0, the electric field contribution from q2 will be 0.
Step 3: Calculate the total electric field at point P:
The total electric field at point P is the vector sum of E1 and E2:
E_total = E1 + E2
Since E2 is 0, the total electric field is:
E_total = E1
Finally, we have the magnitude and direction of the electric field at point P. Note that it is a vector quantity, so we need to represent it in terms of its magnitude and direction.
For each:
Magnitude is E = kq/r^2
Direction do a tan-1.
The effect will be the vector sum of the two vectors (just sum the x and y coordinates of each)