If the 3e charge is at the origin and the Magnesium nucleus is 3cm from the origin in the +x direction, at what position (in centimeters) along the x-axis should an electron be placed so that it (the electron) experiences no net force? x =
E=kq₁/x²=kq₂/(3-x)²
3e/x²=12e/(3-x)²
9-6x+x²=4x²
3x²+6x-9=0
x²+2x-3=0
x=1 cm
Thank you very much
What is the magnitude of the net electrostatic force a Boron nucleus would experience at the point halfway between the Magnesium nucleus and the 3e charge on the y axis in N?
Its not ok. If x=1cm, the e- have a net force to the rigth
x=-3cm
To find the position along the x-axis where an electron experiences no net force, we need to calculate the distance (x) between the electron and the origin.
We know that the charge of the electron is opposite in sign to the charge of the Magnesium nucleus. The charge of an electron is -e, where e is the elementary charge (approximately 1.6 x 10^-19 coulombs). And the charge of the Magnesium nucleus is +3e, where 3e represents three times the elementary charge.
The electric force between two charges is given by Coulomb's law:
F = k * (q1 * q2) / r^2
Where F is the force, k is the Coulomb's constant (approximately equal to 9 x 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
Since the net force experienced by the electron should be zero, we can set up the equation:
F_electron = F_nucleus
k * ((-e) * (3e)) / (x^2) = k * ((3e) * (3e)) / (3 cm)^2
Simplifying the equation:
(-e) * (3e) / (x^2) = (3e) * (3e) / (3^2)
Now, we can solve for x:
(-e) * (3e) * (3^2) = (3e) * (3e) * (x^2)
(-e) * (3e) * 9 = (3e) * (3e) * (x^2)
-27e^2 = 9e^2 * (x^2)
-27 = 9 * (x^2)
x^2 = -3
Taking the square root of both sides of the equation:
x = ±√(-3)
Since the distance cannot be negative, we can disregard the negative solution.
Therefore, the position along the x-axis at which the electron experiences no net force is x = √(-3) cm.