Question

An electron on the axis of an electric dipole is 26 nm from the center of the dipole. What is the magnitude of the electrostatic force on the electron if the dipole moment is 4.1 ✕ 10−29 C · m? Assume that 26 nm is much larger than the dipole charge separation.

Answer 1

`F = 6.72 * 10^(-15) N`

Step-by-step explanation:

As we know that the electric field due to dipole on its axis is given as

`E = (2kP)/(r^3)`

here we know that

`P = 4.1 * 10^(-29) Cm`

`r = 26 nm`

now the electric field is given as

`E = (2(9 * 10^9)(4.1 * 10^(-29)))/((26 * 10^(-9))^3)`

`E = 4.2 * 10^5 N/C`

now we know the force is given as

`F = qE`

`F = 1.6* 10^(-19)(4.2 * 10^5)`

`F = 6.72 * 10^(-15) N`