Question

What is the acceleration of an electron when it is 5.5mm from a thin rod of length 40cm and charge 6microC? Assume the charge is distributed evenly along the length of the rod and that the electron is not near either end of the rod.

Answer 1

Since 5.5mm is much smaller than 40cm, and since the electron is not near an end of the rod, the field experienced by the electron is (to a good approximation) the same as the field from an infinite length rod of the same linear charge density, λ = 6µC/0.4m = 15 µC/m.

The field a distance r from an infinite rod of linear charge density λ is:
E = λ/(2πrε₀)
E = 15x10⁻⁶ /(2π x 5.5x10⁻³ x 8.865x10⁻¹²)
. .= 4.90x10⁷ N/C


Force on electron is:
F = qE
. .= -1.6x10⁻¹⁹ x 4.90x10⁷
. .= -7.84x10⁻¹² N (the – sign just means the direction is radially inwards (towards rod))
______________

Acceleration of electron (using F= ma) is:
a = F/m
. .= -7.84x10⁻¹² / (9.11x10⁻³¹)
. . = -8.61x10¹⁸ m/s² (the – sign just means the direction is radially inwards (towards rod))

The magnitude of the acceleration is therefore 8.61x10¹⁸ m/s²