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kari74 [83]
3 years ago
13

The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 4.50×10−9 m2 , a plate separ

ation of 8.1×10−9 m , and a dielectric with a dielectric constant of 4.6.What is the energy stored in such a cell membrane if the potential difference across it is 7.55×10−2 V
Physics
1 answer:
Marizza181 [45]3 years ago
6 0

The energy stored in the membrane is 6.44\cdot 10^{-14} J

Explanation:

The capacitance of a parallel-plate capacitor is given by

C=\frac{k\epsilon_0 A}{d}

where

k is the dielectric constant of the material

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

For the membrane in this problem, we have

k = 4.6

A=4.50\cdot 10^{-9} m^2

d=8.1\cdot 10^{-9} m

Substituting, we find its capacitance:

C=\frac{(4.6)(8.85\cdot 10^{-12})(4.50\cdot 10^{-9})}{8.1\cdot 10^{-9}}=2.26\cdot 10^{-11} F

Now we can find the energy stored: for a capacitor, it is given by

U=\frac{1}{2}CV^2

where

C=2.26\cdot 10^{-11} F is the capacitance

V=7.55\cdot 10^{-2} V is the potential difference

Substituting,

U=\frac{1}{2}(2.26\cdot 10^{-11} F)(7.55\cdot 10^{-2})^2=6.44\cdot 10^{-14} J

Learn more about capacitors:

brainly.com/question/10427437

brainly.com/question/8892837

brainly.com/question/9617400

#LearnwithBrainly

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