Explanation:
The given data is as follows.
q = 9.0 pC = [tex]9 \times 10^{-12} C[/tex] (as 1 pC = [tex]10^{-12} C[/tex])
t = 0.47 ms = [tex]0.47 \times 10^{-3} s[/tex] (as 1 ms = [tex]10^{-3} sec[/tex])
As we know that relation between charge and current is as follows.
I = [tex]\frac{q}{t}[/tex]
Putting the given values into the above formula as follows.
I = [tex]\frac{q}{t}[/tex]
= [tex]\frac{9 \times 10^{-12} C}{0.47 \times 10^{-3} s}[/tex]
= [tex]19.14 \times 10^{-9} A[/tex]
or, = 19.14 nA
Thus, we can conclude that the average current through the cell membrane is 19.14 nA.
