Charge q1 = 7.00 μC is at the origin, and charge q2 = -5.00 μC is on the x-axis, 0.300 m from
the origin as shown below. (a) Find the magnitude and direction of the electric field at point P,
which has coordinates (0, 0.400) m. (b) Find the force on a charge of 2.00 × 10-8 C placed at
P.
a) The magnitude of the electric field at point P is 8.33 N/C and the direction is in the positive y-direction.
b) The force on a charge of 2.00 × 10-8 C placed at P is 0.0167 N.
To find the magnitude and direction of the electric field at point P, we can use the equation for electric field:
Electric Field (E) = k * Q / r^2
Where:
- k is the Coulomb's constant, approximately 9.0 x 10^9 N·m^2/C^2
- Q is the charge creating the electric field
- r is the distance from the source charge to the point where the electric field is being measured
(a) Let's calculate the magnitude of the electric field at point P:
The distance from the origin to P is the square root of the sum of the x and y components:
r = sqrt((0.300 m)^2 + (0.400 m)^2)
r = sqrt(0.09 m^2 + 0.16 m^2)
r = sqrt(0.25 m^2)
r = 0.5 m
The electric field E created by charge q1 at point P:
E1 = (k * q1) / r^2
E1 = (9.0 x 10^9 N·m^2/C^2 * 7.00 μC) / (0.5 m)^2
E1 = 63 x 10^-3 N/C
The electric field E created by charge q2 at point P:
E2 = (k * q2) / r^2
E2 = (9.0 x 10^9 N·m^2/C^2 * -5.00 μC) / (0.5 m)^2
E2 = -90 x 10^-3 N/C
To find the net electric field at point P, we need to add the electric field vectors:
E_net = E1 + E2
E_net = 63 x 10^-3 N/C + (-90 x 10^-3 N/C)
E_net = -27 x 10^-3 N/C
The magnitude of the electric field at point P is 27 x 10^-3 N/C.
To find the direction, we need to find the angle θ between the x-axis and the net electric field vector.
Using trigonometry:
tan(θ) = (0.400 m) / (0.300 m)
θ = arctan(0.400 m / 0.300 m)
θ ≈ 53.1°
Therefore, the direction of the electric field at point P is approximately 53.1° with respect to the positive x-axis.
(b) To find the force on a charge of 2.00 x 10^-8 C placed at point P, we can use the equation for electric force:
Electric Force (F) = q * E
Where:
- q is the charge experiencing the electric force
- E is the electric field at that point (which we calculated to be -27 x 10^-3 N/C)
F = (2.00 x 10^-8 C) * (-27 x 10^-3 N/C)
F = -54 x 10^-11 N
Therefore, the force on the charge of 2.00 x 10^-8 C placed at point P is approximately -54 x 10^-11 N.
To find the magnitude and direction of the electric field at point P, we can use the principle of superposition.
(a) To find the electric field at point P, we need to calculate the electric field due to each charge separately and then add them up.
1. Electric field due to q1 at point P:
The electric field due to a point charge is given by Coulomb's Law: E = k*q/r^2, where E is the electric field, k is the electrostatic constant (8.99 × 10^9 N·m^2/C^2), q is the charge, and r is the distance between the charge and the point where we want to find the electric field.
In this case, q1 = 7.00 μC = 7.00 × 10^-6 C and r1 = distance between q1 and P = 0.400 m.
Using the formula, E1 = k*q1/r1^2
2. Electric field due to q2 at point P:
Similarly, the electric field due to q2 at point P is given by E2 = k*q2/r2^2, where q2 = -5.00 μC = -5.00 × 10^-6 C and r2 = distance between q2 and P = 0.300 m.
Now, we can simply add the electric fields due to q1 and q2 at point P to get the total electric field at P: E = E1 + E2. Note that we need to consider the signs of the charges while adding.
(b) To find the force on a charge of 2.00 × 10^-8 C placed at point P, we can use the equation F = q * E, where F is the force on the charge, q is the charge, and E is the electric field at that point.
Now we have all the information we need to calculate the answers. Just substitute the values for each step and perform the necessary calculations.