Determine the magnitude of the force between an electric dipole with a dipole moment of 2 ✕ 10-29 C m and an electron. The electron is positioned r = 25nm from the centre of the dipole, along the dipole axis. Hint: Assume that r >> d, with d the charge separation distance in the dipole.
Maybe I'm oversimplifying but this looks like a simple Coulomb's Law with q1 = 2e-29, q2 = 1.67e-19 and r = 25nm.
To determine the magnitude of the force between the electric dipole and the electron, we can use Coulomb's Law. Coulomb's Law states that the force between two charged particles is given by:
F = (k * q1 * q2) / r^2
Where:
F is the force between the particles
k is the electrostatic constant (k ≈ 8.99 x 10^9 N m^2/C^2)
q1 and q2 are the charges of the two particles
r is the distance between them
In the given problem, the electric dipole has a dipole moment of 2 x 10^-29 C m, which is equivalent to the product of charge (q) and the distance of separation (d) between the charges in the dipole (q * d). Therefore, we have:
q * d = 2 x 10^-29 C m
From the hint, it is mentioned that r >> d. This means that the distance between the electron and the center of the dipole is much larger than the separation distance in the dipole.
So, we can approximate the force between the dipole and the electron as the force between a point charge (the dipole) and another point charge (the electron). Using this approximation, the charge of the dipole can be assumed to be split into two equal and opposite charges (+q and -q), with a separation distance equal to d.
Hence, q = (1/2)*(2 x 10^-29 C) = 1 x 10^-29 C (charge of either +q or -q)
Now, we can calculate the force between the dipole and the electron:
F = (k * q1 * q2) / r^2
= (8.99 x 10^9 N m^2/C^2) * (1 x 10^-29 C) * (1.6 x 10^-19 C) / (25 x 10^-9 m)^2
= (1.4384 x 10^-9 N m^2/C^2) / 6.25 x 10^-18 m^2
= 2.31 x 10^8 N
Therefore, the magnitude of the force between the electric dipole and the electron is approximately 2.31 x 10^8 Newtons.