A typical cell has a membrane potential of -70 , meaning that the potential inside the cell is 70 less than the potential outside due to a layer of negative charge on the inner surface of the cell wall and a layer of positive charge on the outer surface. This effectively makes the cell wall a charged capacitor. Because a cell's diameter is much larger than the wall thickness, it is reasonable to ignore the curvature of the cell and think of it as a parallel-plate capacitor. How much energy is stored in the electric field of a 50--diameter cell with a 7.0--thick cell wall whose dielectric constant is 9.0?

energy is stored is 2.2 × 10⁻¹³ J

Explanation:

The capacitance  of the cell is given with the expression

C = (KE₀A) / d

k is the dielectric constant, A is the area of the cell, d is the thickness of the cell.

Now given that; the diameter is 50,

Area A = 4πR²

A = 4π × ( 25 × 10⁻⁶ m)²

A = 7850×10⁻¹² m²

our capacitance C = (KE₀A) / d

C = [9 ( 8.85 × 10⁻¹² C²/N.m²  × 7850×10⁻¹² m² )] / 7×10⁻⁹ m

C = 8.93 × 10⁻¹¹ F

Now Energy stored

E = 1/2 × CV²

E = 1/2 × (8.93 × 10⁻¹¹ F) × ( 70 × 10⁻³ V)²

E = 2.2 × 10⁻¹³ J

Therefore energy is stored is 2.2 × 10⁻¹³ J

if they have a finite resultant

energy can be converted from one form to another, but it cannot be created or destroyed in ordinary chemical or physical means. do not understand what you are asking but hope this