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Chapter 17: Problem 95

The potential difference across a cell membrane is \(-90 \mathrm{mV} .\) If the membrane's thickness is \(10 \mathrm{nm},\) what is the magnitude of the electric field in the membrane? Assume the field is uniform.

Short Answer

Expert verified
Answer: The magnitude of the electric field in the cell membrane is \(9 \times 10^{6} \, \frac{\mathrm{V}}{\mathrm{m}}\).

Step by step solution

01

Convert the units to the base units

First, we need to convert the given values of potential difference and distance into base units. For potential difference, \(1 \, \mathrm{mV} = 1 \times 10^{-3} \, \mathrm{V}\). For distance, \(1 \, \mathrm{nm} = 1 \times 10^{-9} \, \mathrm{m}\). Let's do these conversions: \(V = -90 \, \mathrm{mV} = -90 \times 10^{-3} \, \mathrm{V} = -0.090 \, \mathrm{V}\) \(d = 10 \, \mathrm{nm} = 10 \times 10^{-9} \, \mathrm{m} = 1 \times 10^{-8} \, \mathrm{m}\)
02

Calculate the magnitude of the electric field in the membrane

Now that we have converted our values, we can use the electric field equation to find the magnitude of the electric field in the membrane. We will use the absolute values for both \(V\) and \(d\) since we only want the magnitude (and not the direction) of the electric field. \(E = \cfrac{|V|}{|d|} = \cfrac{0.090 \, \mathrm{V}}{1 \times 10^{-8} \, \mathrm{m}}\) \(E = 9 \times 10^{6} \, \frac{\mathrm{V}}{\mathrm{m}}\) The magnitude of the electric field in the cell membrane is \(9 \times 10^{6} \, \frac{\mathrm{V}}{\mathrm{m}}\).

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Most popular questions from this chapter

An alpha particle (charge \(+2 e\) ) moves through a potential difference \(\Delta V=-0.50 \mathrm{kV} .\) Its initial kinetic energy is $1.20 \times 10^{-16} \mathrm{J} .$ What is its final kinetic energy?
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