The physics of electric field between two charges.
What is the magnitude and direction of the electric field at a point midway between a
-8.0 ìC charge and a +6.0 ìC charge? The -8.0 ìC charge is 4.0 cm to the left of the +6.0 ìC charge.
If a charge of -2.0 ìC was placed at the midway point, what force (magnitude and direction) would act on the charge?
First we find the electric field produced by each charge at the midway point.
The electric field produced by a point charge at a distance r from the charge is given by the formula:
E = k*Q/r^2
where E is the electric field, k is Coulomb's constant (8.99 * 10^9 N*m^2/C^2), Q is the charge, and r is the distance from the charge.
1. Electric field produced by the -8.0 μC charge:
E1 = k*Q1/r^2
E1 = (8.99 * 10^9 N*m^2/C^2) * (-8.0 * 10^(-6) C) / (0.02 m)^2
E1 = (-1.4376 * 10^5 N/C) to the right
2. Electric field produced by the +6.0 μC charge:
E2 = k*Q2/r^2
E2 = (8.99 * 10^9 N*m^2/C^2) * (6.0 * 10^(-6) C) / (0.02 m)^2
E2 = (1.0782 * 10^5 N/C) to the left
Now we find the net electric field at the midway point by adding the two electric fields:
Net electric field (E_net) = E1 + E2
E_net = -1.4376 * 10^5 N/C to the right + 1.0782 * 10^5 N/C to the left
E_net = (-1.4376 + 1.0782) * 10^5 N/C
E_net = -0.3594 * 10^5 N/C
So the magnitude of the electric field at the midway point is 0.3594 * 10^5 N/C, and its direction is to the left.
Now we find the force acting on the -2.0 μC charge placed at the midway point. The force on a charge in an electric field is given by the formula:
F = q*E
where F is the force, q is the charge, and E is the electric field.
Force on -2.0 μC charge (F) = q*E_net
F = (-2.0 * 10^(-6) C) * (-0.3594 * 10^5 N/C)
F = (0.7188 N) to the right
Therefore, the force acting on the -2.0 μC charge placed at the midway point has a magnitude of 0.7188 N and its direction is to the right.
To determine the magnitude and direction of the electric field at a point between two charges, we can use the principle of superposition. The electric field at that point will be the vector sum of the electric fields created by each charge individually.
In this scenario, there are two charges: -8.0 ìC and +6.0 ìC. Let's denote the -8.0 ìC charge as Q1 and the +6.0 ìC charge as Q2.
First, we need to calculate the electric field generated by each charge separately using Coulomb's Law:
Electric field due to Q1 (E1):
The electric field (E) produced by a point charge (Q) at a distance (r) from it is given by the equation:
E = k * |Q| / r^2, where k is the electrostatic constant (k = 9 × 10^9 N m^2/C^2).
Given:
Q1 = -8.0 ìC
r1 = 4.0 cm = 0.04 m (since 1 cm = 0.01 m)
Using the equation, we can calculate E1:
E1 = (9 × 10^9 N m^2/C^2) * (8.0 × 10^-6 C) / (0.04 m)^2
Simplifying the expression, we find that E1 ≈ 45,000 N/C directed rightwards.
Electric field due to Q2 (E2):
Given:
Q2 = +6.0 ìC
r2 = 4.0 cm = 0.04 m
Using the same equation, we can calculate E2:
E2 = (9 × 10^9 N m^2/C^2) * (6.0 × 10^-6 C) / (0.04 m)^2
Simplifying the expression, we find that E2 ≈ 34,000 N/C directed leftwards.
Now, to find the electric field at the midway point between the charges (let's call it P), we need to sum the individual electric fields:
Electric field at point P:
The electric field at point P (E_total) can be found by vector addition of E1 and E2:
E_total = E1 + E2
Given that E1 is rightwards and E2 is leftwards, the electric fields will partially cancel each other, resulting in the net electric field at P.
Magnitude of E_total:
|E_total| = |E1| + |E2| = 45,000 N/C + 34,000 N/C = 79,000 N/C
Direction of E_total:
Since the magnitudes of E1 and E2 are different, the electric field at point P will not completely cancel out. However, since E1 has a larger magnitude, the net electric field at P will be directed rightwards.
Now, let's move on to the second part of your question:
If a charge of -2.0 ìC is placed at the midway point between the two charges, we need to determine the force acting on this charge.
The force experienced by a charge (Q) in an electric field (E) is given by the equation:
F = |Q| * |E|
Given:
Q = -2.0 ìC
|E_total| = 79,000 N/C (from the previous calculations)
Substituting the values, we find that:
F = (2.0 × 10^-6 C) * (79,000 N/C)
Simplifying the expression, we see that the magnitude of the force acting on the charge is F ≈ 158 N.
As for the direction, since the electric field at the midway point is directed rightwards (as calculated earlier), the direction of the force on the charge will be the same, i.e., the force will be directed rightwards.
Therefore, the force acting on the charge is approximately 158 N directed rightwards.