The energy stored in the membrane is ![6.44\cdot 10^{-14} J](https://tex.z-dn.net/?f=6.44%5Ccdot%2010%5E%7B-14%7D%20J)
Explanation:
The capacitance of a parallel-plate capacitor is given by
![C=\frac{k\epsilon_0 A}{d}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7Bk%5Cepsilon_0%20A%7D%7Bd%7D)
where
k is the dielectric constant of the material
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](https://tex.z-dn.net/?f=A%3D4.50%5Ccdot%2010%5E%7B-9%7D%20m%5E2)
![d=8.1\cdot 10^{-9} m](https://tex.z-dn.net/?f=d%3D8.1%5Ccdot%2010%5E%7B-9%7D%20m)
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](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B%284.6%29%288.85%5Ccdot%2010%5E%7B-12%7D%29%284.50%5Ccdot%2010%5E%7B-9%7D%29%7D%7B8.1%5Ccdot%2010%5E%7B-9%7D%7D%3D2.26%5Ccdot%2010%5E%7B-11%7D%20F)
Now we can find the energy stored: for a capacitor, it is given by
![U=\frac{1}{2}CV^2](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B2%7DCV%5E2)
where
is the capacitance
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](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B1%7D%7B2%7D%282.26%5Ccdot%2010%5E%7B-11%7D%20F%29%287.55%5Ccdot%2010%5E%7B-2%7D%29%5E2%3D6.44%5Ccdot%2010%5E%7B-14%7D%20J)
Learn more about capacitors:
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brainly.com/question/8892837
brainly.com/question/9617400
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