Question

1. A bacterium has 1.8 × 1016 protons and a net charge of 9.4 pC.

If you paired them up, what fraction of the protons would have no electrons?

2. Common transparent adhesive tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece’s weight.

Assuming equal point charges as an approximation, calculate the magnitude of each charge, in coulombs, if the electrostatic force is great enough to support the weight of a 0.65 mg piece of tape held 1.1 cm above another.

Answer 1

1.

Number of protons, N = 1.8 x 10¹⁶

Net charge, q = 9.4 pC = 9.4 x 10^-12 C

If n is the number of elementary particles corresponding to a charge q then

q = ne

n = q/e = 9.4 x 10^-12 /1.6 x 10^-19 = 5.875 x 10⁷

charge corresponds to N protons, Q = Ne = 1.8 x 10¹⁶ x 1.6 x 10^-19 =1.125 x 10³ C

fraction of protons that would have no electrons when paired = (N – n)/n x 100 = (N/n – 1) x100

                                = (1.8 x 10¹⁶/5.875 x 10⁷ -1) x 100 = 3.06 x 10¹¹%

2.

Mass of each top, m = 0.65 mg = 0.65 x 10^-6kg

Vertical distance between two tapes, d = 1.1 cm = 0.011 m

Let q be the charge on each tape such that the electric force of repulsion between the two tapes balances the weight of the upper tape.

Electric force = weight

1/4πε₀ q x q/d² = mg

9 x 10⁹ x qxq/0.011 x 0.011 = 0.65 x 10^-6 x 9.8

Charge on each tape, q = 0.292 C