Question

Capacitors can be used to count individual electrons, if adding a single electron to one plate causes a measurable voltage increase between the plates. (a) If the capacitor has capacitance C and there are N electrons on one plate (and N missing from the other), write a formula for the voltage between the plates. (b) If a single extra electron is added to a plate (N → N + 1 or ∆N = 1), by how much does the voltage between the plates change? Does the amount by which the voltage changes depend on the number of electrons that were already separated by the plates? (c) If the smallest voltage change you can reliably measure is 1.0 mV, what value of C is required to notice the arrival of a single electron? (d) A small off-the-shelf capacitor may have a capacitance of 1.0 nF. What would be the minimum number of electrons moving from one plate to the other than would result in a measurable voltage change? (e) It is possible to fabricate micron-scale capacitors with parallel conducting square plates separated by a d = 100 nm thick insulating oxide layer with a dielectric constant of κ = 3.0. How long should each side of the square plates be to have the capacitance required in (c) to measure individual electrons? Based on your result, does the “parallel-plate formula” apply here? Explain.

Answer 1

a)

C = capacitance

N = number of electrons

e = charge on single electron

Q = total charge stored by the capacitor = Ne

V = Voltage across the plates

using the equation

Q = CV

V = Q/C

V = Ne/C

b)

after adding the electron, Voltage is given as

V' = (N + 1) e/C

Potential difference is given as

V' - V = (N + 1 - N) (e/C)

V = e/C

no the potential difference does not depend on the number of electrons already separated

c)

V = 0.001 volts

e = 1.6 x 10^-19 C

hence

0.001 = (1.6 x 10^-19)/C

C = 1.6 x 10^-16 F