Answer :
This electric field is much lower than the maximum allowed electric field of 6.0 x 10^-8 V/m, so the thickness of the trace that you have chosen is appropriate.
To determine the thickness of the gold trace, you will need to use the equation for resistance of a wire:
R = (resistivity of gold) / (cross-sectional area of trace)
where R is the resistance of the trace, resistivity of gold is the inherent resistance of gold, and cross-sectional area of trace is the cross-sectional area of the trace.
You can find the resistivity of gold from a table of electrical conductivity values. At room temperature, the resistivity of gold is about 2.44 x 10^-8 Ω*m.
Next, you need to determine the cross-sectional area of the trace. Since the width of the trace is fixed at 2 mm, and the distance between the devices is 15 mm, the cross-sectional area of the trace will be a rectangle with dimensions of 2 mm by 15 mm, or 30 mm^2.
Now that you have the resistivity of gold and the cross-sectional area of the trace, you can plug these values into the resistance equation to find the resistance of the trace:
R = (2.44 x 10^-8 Ω*m) / (30 mm^2) = 8.13 x 10^-10 Ω
Next, you can use Ohm's law to find the current flowing through the trace:
I = V / R
where I is the current, V is the voltage, and R is the resistance of the trace.
Since the current flowing through the trace needs to be 2.5 μA, and the resistance of the trace is 8.13 x 10^-10 Ω, you can solve for the voltage:
V = I * R = (2.5 x 10^-6 A) * (8.13 x 10^-10 Ω) = 2.03 x 10^-15 V
Finally, you can use the electric field equation to find the electric field between the devices:
E = V / d
where E is the electric field, V is the voltage, and d is the distance between the devices.
Since the distance between the devices is 15 mm, and the voltage between the devices is 2.03 x 10^-15 V, the electric field between the devices is:
E = (2.03 x 10^-15 V) / (15 mm) = 1.35 x 10^-16 V/m
Learn more about electricity, here https://brainly.com/question/8971780
#SPJ4
