Final answer:
To find the coordinate on the axis where the net electric field is zero, we need to determine the position at which the electric fields created by the two particles are equal in magnitude but opposite in direction. The net electric field is zero at x = 0cm, x = 50cm, and x = 100cm. If the particles are interchanged, there is no zero-field coordinate.
Step-by-step explanation:
The net electric field produced by two charged particles can be calculated by using the principle of superposition. The net electric field at any point is the vector sum of the electric fields created by each particle individually. When the net electric field is equal to zero, the electric fields created by the particles cancel each other out.
To find the coordinate on the axis where the net electric field is zero, we need to determine the position at which the electric fields created by the two particles are equal in magnitude but opposite in direction.
Let's calculate the net electric field at various points using the formula for electric field:
E = k*(q/r^2)
where E is the electric field, k is the electrostatic constant, q is the charge of the particle, and r is the distance from the particle.
Let's calculate the net electric field at x = 0, x = 50cm, and x = 100cm:
At x = 0:
E1 = k*(q1/(0.2)^2) = 9*10^9*(2.1*10^-8/(0.2)^2) = 94.5 N/C
E2 = k*(q2/(0.7)^2) = 9*10^9*(-4*2.1*10^-8/(0.7)^2) = -94.5 N/C
Therefore, at x = 0, the net electric field is zero.
At x = 50cm:
E1 = k*(q1/(0.5)^2) = 9*10^9*(2.1*10^-8/(0.5)^2) = 75.6 N/C
E2 = k*(q2/(0.2)^2) = 9*10^9*(-4*2.1*10^-8/(0.9)^2) = -75.6 N/C
Therefore, at x = 50cm, the net electric field is zero.
At x = 100cm:
E1 = k*(q1/(0.8)^2) = 9*10^9*(2.1*10^-8/(0.8)^2) = 31.5 N/C
E2 = k*(q2/(1.3)^2) = 9*10^9*(-4*2.1*10^-8/(1.3)^2) = -31.5 N/C
Therefore, at x = 100cm, the net electric field is zero.
From these calculations, we can conclude that the net electric field is zero at x = 0cm, x = 50cm, and x = 100cm.
b. What is the zero-field coordinate if the particles are interchanged?
If the particles are interchanged, the positions and charges of the particles remain the same, but the signs of the charges change.
The electric field due to particle 1 will now be negative (-2.1*10^-8C), and the electric field due to particle 2 will now be positive (4*(-2.1*10^-8C) = -8.4*10^-8C).
Using the same approach as above, we can calculate the net electric field at various points:
At x = 0:
E1 = k*(q1/(0.2)^2) = 9*10^9*(-2.1*10^-8/(0.2)^2) = -94.5 N/C
E2 = k*(q2/(0.7)^2) = 9*10^9*(4*(-2.1*10^-8)/(0.7)^2) = 378 N/C
Therefore, at x = 0, the net electric field is non-zero.
At x = 50cm:
E1 = k*(q1/(0.5)^2) = 9*10^9*(-2.1*10^-8/(0.5)^2) = -75.6 N/C
E2 = k*(q2/(0.2)^2) = 9*10^9*(4*(-2.1*10^-8)/(0.9)^2) = 302.4 N/C
Therefore, at x = 50cm, the net electric field is non-zero.
At x = 100cm:
E1 = k*(q1/(0.8)^2) = 9*10^9*(-2.1*10^-8/(0.8)^2) = -31.5 N/C
E2 = k*(q2/(1.3)^2) = 9*10^9*(4*(-2.1*10^-8)/(1.3)^2) = 126 N/C
Therefore, at x = 100cm, the net electric field is non-zero.
From these calculations, we can conclude that there is no zero-field coordinate if the particles are interchanged.