Answer
(a) $7.2\times 10^{-13}A$
(b) decreases by a factor of 2.
Work Step by Step
We know that
$R=\frac{\rho L}{A}$
We plug in the known values to obtain:
$R=\frac{1.3\times 10^7\times 8\times 10^{-9}}{10^{-12}}$
$R=1.04\times 10^{11}\Omega$
Now according to Ohm's law
$V=IR$
We plug in the known values to obtain:
$0.75\times 10^{-3}=I\times 1.04\times 10^{11}$
This simplifies to:
$I=7.2\times 10^{-13}A$
(b) If the thickness of the membrane is doubled then the resistance will also be doubled because resistance is directly proportional to thickness. We know that $R=\frac{V}{I}$ . This equation shows that resistance and current are inversely proportional, so the current will decrease by a factor of 2 when the resistance is doubled.
