Electrical Energy Storage?

Question

The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 4.75 x 10^-9 m^2, a plate separation of 8.5 x 10^-9 m, and a dielectric with a dielectric constant of 4.5. (a) What is the energy stored in such a cell membrane if the potential difference across it is 0.0725 V? (b) Would your answer to part (a) increase, decrease, or stay the same if the thickness of the cell membrane is increased? Explain.

Thanks so much!

Answer 1

Capacitance
C= e A/d
or C=Q/V
Energy stored in the capacitor
W=0.5CV^2

a) This a straight forward problem
W=0.5 e A/d (V^2)
W=0.5 x 4.5 x 4.75 x 10^-9 m^2 / 8.5 x 10^-9 m=
W=...

b) since d is the thickness and the energy is inversely proportional to energy stored in the capasitor then with the increase of the thickness of the membrane the energy will decrease.

Answer 2

Compute the vacuum capacitance = e*A/d where A is the membrane area and d is the membrance separation, and where e is 8.9 pF/m. 1 pF = 1*10^-12 farads.

Since the dielectric const. is 4.5, multiply the above answer by 4.5 to get the correct capacitance.

Then compute the energy = (1/2)*C*V^2 where C is the capacitance, and V is the voltage.

Answer 3

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