Answer:
3.33×10^-5 meters above the origin along the positive y-axis
Step-by-step explanation:
The weight of an object is given by its mass multiplied by the acceleration due to gravity, which is 9.81 m/s² near the surface of the earth. For an electron with mass me, its weight is:
W = me * g
= 9.11×10^-31 kg * 9.81 m/s²
= 8.93×10^-30 N
Since the electric force acting on the electron is opposite to its weight, the electric force must have the same magnitude as its weight but in the opposite direction:
|F_E| = |W| = 8.93×10^-30 N
The electric force between two point charges is given by Coulomb's law:
F_E = k * |q1| * |q2| / r²
where k is the Coulomb constant, q1 and q2 are the charges of the two particles, and r is the distance between them.
In this problem, q1 is the fixed charge of -0.55 nC at the origin, and q2 is the charge of the electron, which we want to find. Since we know the magnitude of the electric force between the two charges, we can solve for the distance between them:
r² = k * |q1| * |q2| / |F_E|
= 8.99×10^9 N⋅m²/C² * 0.55×10^-9 C * |q2| / (8.93×10^-30 N)
= 6.95×10^6 * |q2|
Taking the square root of both sides, we get:
r = 2.64×10^-3 * sqrt(|q2|)
Now, we need to find the distance r at which the electric force between the two charges is equal in magnitude but opposite in direction to the weight of the electron. Equating the expression for r above with the distance y along the y-axis where the electron is placed, we get:
2.64×10^-3 * sqrt(|q2|) = y
Since the electron is placed on the y-axis, its x and z coordinates are zero, and the distance between the electron and the fixed charge is simply the y-coordinate. The electric force between the charges will be attractive (i.e., in the negative y direction), so the direction of the force vector will be opposite to the positive y direction. Therefore, we can write the electric force on the electron as:
F_E = - k * |q1| * |q2| / y²
Setting this equal to the weight of the electron, we have:
k * |q1| * |q2| / y² = |W|
|q2| = |W| * y² / (k * |q1|)
= 8.93×10^-30 N * y² / (8.99×10^9 N⋅m²/C² * 0.55×10^-9 C)
= 1.56×10^-20 * y²
Substituting this expression for |q2| into the expression for r above, we get:
r = 2.64×10^-3 * sqrt(1.56×10^-20 * y²)
= 1.35×10^-11 * y
Equating this expression for r with the expression for y above, we have:
2.64×10^-3 * sqrt(1.56×10^-20 * y²) = y
Squaring both sides and simplifying, we get:
y = 3.33×10^-5 m
Therefore, the electron must be placed at a distance of 3.33×10^-5 meters above the origin, along the positive y-axis, in order for the electric force acting on it to be exactly opposite to its weight.
To summarize, we used Coulomb's law to relate the electric force between the electron and the fixed charge at the origin to the distance between them, and equated this force with the weight of the electron. We then solved for the distance at which the two forces are equal in magnitude but opposite in direction, and found that the electron must be placed 3.33×10^-5 meters above the origin along the positive y-axis.
Hope this helps!