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 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?
1 answer
1 answer