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

Answer:

$F = 6.72 \times 10^{-15} N$

Explanation:

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

$E = \frac{2kP}{r^3}$

here we know that

$P = 4.1 \times 10^{-29} Cm$

$r = 26 nm$

now the electric field is given as

$E = \frac{2(9 \times 10^9)(4.1 \times 10^{-29})}{(26 \times 10^{-9})^3}$

$E = 4.2 \times 10^5 N/C$

now we know the force is given as

$F = qE$

$F = 1.6\times 10^{-19}(4.2 \times 10^5)$

$F = 6.72 \times 10^{-15} N$