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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?

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Answer:

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

The energy that should be stored in the electric field should be 2.2 × 10⁻¹³ J.

Calculation of the energy:

Since

The capacitance  of the cell should be

C = (KE₀A) / d

Here,

k is the dielectric constant,

A is the area of the cell,

d is the thickness of the cell.

Now the diameter is 50,

So,

Area A = 4πR²

A = 4π × ( 25 × 10⁻⁶ m)²

A = 7850×10⁻¹² m²

Now

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 should be

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.

Learn more about cell here: https://brainly.com/question/22268502

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