The energy stored in the membrane is 
Explanation:
The capacitance of a parallel-plate capacitor is given by

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


Substituting, we find its capacitance:

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

where
is the capacitance
is the potential difference
Substituting,

Learn more about capacitors:
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