Question

A 0.15g plastic bead is charged by the addition of 1.0×1010 excess electrons, What electric field E⃗ (strength) will cause the bead to hang suspended in the air? What electric field E⃗ (direction) will cause the bead to hang suspended in the air, up or down?

Answer 1

The magnitude of the electric field for the bead to hang suspended in air is 9.19×10⁵V/m and the field is directed downwards.

If n electrons each of charge e are added to the bead, the total charge q on the bead is given by

$q=ne$

Substitute 1.0×10¹⁰ for n and 1.6×10⁻¹⁹C for e.

$q=ne\\ =(1.0*10^1^0)(1.6*10^-^1^9C)\\ =1.6*10^-^9C$

The bead ha a negative charge of magnitude 1.6×10⁻⁹C.

The bead has a mass m and its weight mg is directed downwards. The bead can be balanced in an electric field of strength E if the force it experiences due to the electric field is directed upwards and is equal to the magnitude of its weight. This is shown in the diagram attached.

For equilibrium,

$mg=qE$

Rewrite the expression for E.

$E=\frac{mg}{q}$

Substitute 0.15 g = 0.15×10⁻³kg for m, 9.8 m/s² for g and 1.6×10⁻⁹C for q.

$E=\frac{mg}{q}\\ =\frac{(0.15*10^-^3kg)(9.8m/s^2)}{(1.6*10^-^9C)} \\ =9.1875*10^5V/m$

The magnitude of the electric field required to balance the bead is 9.19×10⁵V/m.

The bead is negatively charged. For the bead to experience a force in the upward direction, the electric field is to be directed down wards.

The electric field is directed downwards.