Answer:
Part a: The current is 1.49x10⁻¹² A
Part b: The current is decreased by a factor of 4.
Explanation:
Part a
The area is given as
A =(1.4*10⁻⁶)² =1.96*10⁻¹² m²
The resistance is given as where resistivity of the membrane material is 1.30 x 10⁷ ohms*m
R=pL/A =(1.3*10⁷)(7.5*10⁻⁹)/(1.96*10⁻¹²)
R=4.97x10¹⁰ ohms
So the resistance is 4.97x10¹⁰ ohms.
I=V/R =86.0x10⁻³/5.77x10¹⁰
I=1.49x10^⁻¹² A
So the current is 1.49x10⁻¹² A
b)
S=So/2=1.4/2 =0.7μm
A=(0.7*10^-6)^2=4.9*10⁻¹³ m²
R=pL/A =(1.3*10⁷)(7.5*10⁻⁹)/(4.9*10⁻¹³)
R=1.98x10¹¹ ohms
So the resistance is 1.98x10¹¹ ohms.
I=V/R =86.0x10⁻³/1.98x10¹¹
I2=4.52x10^⁻¹³ A
So the ratio of the new current vs the old current is as
I2/I=4.52x10^⁻¹³/1.49x10⁻¹²=0.25
So the current is decreased by a factor of 4.
