Question
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?
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?
Answers
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.
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.