Question

Chris is making a capacitor with capacitance of 1 microfarad. The voltage across the plates is 9 volts. How many electrons need to be on the negatively charged plate? The charge of an electron is 1.6 × 10–19 coulombs.

5.6 × 10 ^13 electrons
1.4 × 10 ^-24 electrons
9.0 × 10 ^-6 electrons
1.4 × 10 ^-18 electrons

Answer 1

Capacitance is given by
$C= \frac{Q}{V}$
where Q is the total charge on a plate and V is the voltage across the plates.  Here, Q is the charge on the electron times the number of electrons, or
$C= \frac{Nq}{9.0V} =1 \mu F$
Solving for N, the total number of electrons:
$N= \frac{1.0E-6F \times 9.0V} {1.6E-19C}=5.6 \times 10^{13}$

Answer 2

Explanation :

Capacitance of a capacitor, $C=1\ \mu F=10^{-6}\ F$

Potential across the plates, $V=9\ V$  

We know that the capacitance of a capacitor is given by :

$q=CV$

We have to find the number of electrons.

since, q = ne (Charge is quantised)

n is the number of electrons.

So, ne = CV

$n=\dfrac{CV}{e}$

$n=\dfrac{10^{-6}\ F\times 9\ V}{1.6\times 10^{-19}\ C}$

$n=5.625\times 10^{-13}$

or

$n=5.6\times 10^{-13}$

So, the no. of electrons need to be on the negatively charged plate is $5.625\times 10^{-13}$.

Hence, this is the required solution.