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Chapter 17: Problem 72

In capacitive electrostimulation, electrodes are placed on opposite sides of a limb. A potential difference is applied to the electrodes, which is believed to be beneficial in treating bone defects and breaks. If the capacitance is measured to be \(0.59 \mathrm{pF},\) the electrodes are \(4.0 \mathrm{cm}^{2}\) in area, and the limb is \(3.0 \mathrm{cm}\) in diameter, what is the (average) dielectric constant of the tissue in the limb?

Short Answer

Expert verified
Answer: The average dielectric constant of the tissue in the limb is approximately 1.98.

Step by step solution

01

Write down the capacitance formula in terms of dielectric constant.

We will use the given capacitance formula involving dielectric constant (K), area (A), and distance (d) between the electrodes: $$C = \frac{\epsilon_0 K A}{d}$$ Where \(C\) is the capacitance, \(\epsilon_0\) is the vacuum permittivity, \(K\) is the dielectric constant, \(A\) is the electrode area, and \(d\) is the distance between the electrodes. In this problem, we are trying to solve for dielectric constant \(K\).
02

Substitute the given values into the formula.

Now we substitute the given values into the capacitance formula: Capacitance, \(C = 0.59 \, \mathrm{pF} = 0.59 \times 10^{-12} \, \mathrm{F}\) Area, \(A = 4.0 \, \mathrm{cm^2} = 4.0 \times 10^{-4} \, \mathrm{m^2}\) Diameter, \(d = 3.0 \, \mathrm{cm} = 0.03 \, \mathrm{m}\) Vacuum permittivity, \(\epsilon_0 = 8.854 \times 10^{-12} \, \mathrm{F/m}\)
03

Rearrange the capacitance formula for dielectric constant K.

Now we rearrange the capacitance formula to find the dielectric constant: $$K = \frac{C d}{\epsilon_0 A}$$
04

Calculate the dielectric constant K using the given values.

Substitute the given values into the rearranged capacitance formula to find the dielectric constant: $$K = \frac{(0.59 \times 10^{-12} \, \mathrm{F})(0.03 \, \mathrm{m})}{(8.854 \times 10^{-12} \, \mathrm{F/m})(4.0 \times 10^{-4} \, \mathrm{m^2})}$$
05

Simplify the expression and find the dielectric constant K.

Now we can simplify the expression and find the dielectric constant: $$K \approx 1.98$$ The average dielectric constant of the tissue in the limb is approximately 1.98.

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