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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

A 4.00 - \(\mu \mathrm{F}\) air gap capacitor is connected to a \(100.0-\mathrm{V}\) battery until the capacitor is fully charged. The battery is removed and then a dielectric of dielectric constant 6.0 is inserted between the plates without allowing any charge to leak off the plates. (a) Find the energy stored in the capacitor before and after the dielectric is inserted. [Hint: First find the new capacitance and potential difference.] (b) Does an external agent have to do positive work to insert the dielectric or to remove the dielectric? Explain.
A defibrillator is used to restart a person's heart after it stops beating. Energy is delivered to the heart by discharging a capacitor through the body tissues near the heart. If the capacitance of the defibrillator is $9 \mu \mathrm{F}\( and the energy delivered is to be \)300 \mathrm{J},$ to what potential difference must the capacitor be charged?
Find the electric potential energy for the following array of charges: charge \(q_{1}=+4.0 \mu \mathrm{C}\) is located at \((x, y)=(0.0,0.0) \mathrm{m} ;\) charge \(q_{2}=+3.0 \mu \mathrm{C}\) is located at \((4.0,3.0) \mathrm{m} ;\) and charge \(q_{3}=-1.0 \mu \mathrm{C}\) is located at (0.0,3.0) \(\mathrm{m}\)
It has only been fairly recently that 1.0 -F capacitors have been readily available. A typical 1.0 -F capacitor can withstand up to \(5.00 \mathrm{V}\). To get an idea why it isn't easy to make a 1.0 -F capacitor, imagine making a \(1.0-\mathrm{F}\) parallel plate capacitor using titanium dioxide \((\kappa=90.0\) breakdown strength \(4.00 \mathrm{kV} / \mathrm{mm}\) ) as the dielectric. (a) Find the minimum thickness of the titanium dioxide such that the capacitor can withstand \(5.00 \mathrm{V}\). (b) Find the area of the plates so that the capacitance is \(1.0 \mathrm{F}\)
The inside of a cell membrane is at a potential of \(90.0 \mathrm{mV}\) lower than the outside. How much work does the electric field do when a sodium ion (Na') with a charge of \(+e\) moves through the membrane from outside to inside?
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