Question

A rectangular wafer of pure silicon, with resistivity \(\rho=2300 \Omega \mathrm{m}\) measures \(2.00 \mathrm{~cm}\) by \(3.00 \mathrm{~cm}\) by \(0.0100 \mathrm{~cm}\). Find the maximum resistance of this rectangular wafer between any two faces.

Short answer

Answer: The maximum resistance between any two faces of the silicon wafer is 3,450,000 Ω.

Step-by-step solution

Convert dimensions to meters

First, we need to convert the dimensions of the wafer from centimeters to meters, as the resistivity \(\rho\) is given in \(\Omega\mathrm{m}\). Length: \(2.00~\mathrm{cm} \times \frac{1~\mathrm{m}}{100~\mathrm{cm}} = 0.0200~\mathrm{m}\) Width: \(3.00~\mathrm{cm} \times \frac{1~\mathrm{m}}{100~\mathrm{cm}} = 0.0300~\mathrm{m}\) Thickness: \(0.0100~\mathrm{cm} \times \frac{1~\mathrm{m}}{100~\mathrm{cm}} = 0.000100~\mathrm{m}\)

Calculate resistance for each configuration

Now, we will calculate the resistance for all three possible configurations of the wafer, with different cross-sectional areas. Configuration 1 (Length × Width): \(R_1 = \frac{\rho L}{A} = \frac{2300 \times 0.000100}{0.0200 \times 0.0300} = 383.33~\Omega\) Configuration 2 (Length × Thickness): \(R_2 = \frac{\rho L}{A} = \frac{2300 \times 0.0300}{0.0200 \times 0.000100} = 3450000~\Omega\) Configuration 3 (Width × Thickness): \(R_3 = \frac{\rho L}{A} = \frac{2300 \times 0.0200}{0.0300 \times 0.000100} = 1533333.33~\Omega\)

Find the maximum resistance

Finally, we will compare the three resistance values and choose the maximum value as the solution. Maximum resistance: \(R_\mathrm{max} = \max\{R_1, R_2, R_3\} = \max\{383.33, 3450000, 1533333.33\} = 3450000~\Omega\) Hence, the maximum resistance between any two faces of the silicon wafer is \(3,450,000~\Omega\).