Answer:
[tex]5\cdot 10^{-9} F[/tex]
Explanation:
The capacitance of the electrode is given by:
[tex]C=\frac{Q}{V}[/tex]
where
C is the capacitance
Q is the charge on the electrode
V is the potential difference
In this problem, we have
[tex]Q=45 nC=45\cdot 10^{-9} C[/tex]
V = 9.0 V
Substituting into the equation, we find
[tex]C=\frac{45\cdot 10^{-9}C}{9.0 V}=5\cdot 10^{-9} F[/tex] (5 nF)
