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Some cell walls in the human body have a layer of negative charge on the inside surface and a layer of positive charge of equal magnitude on the outside surface. Suppose that the charge density on either surface is {0.50 10-3 C/m2, the cell wall is 5.0 nm thick, and the cell-wall material is air. (a) Find the magnitude of Ein the wall between the two layers of charge. (b) Find the potential difference between the inside and the outside of the cell. Which is at the higher potential? (c) A typical cell in the human body has a volume of 10-16 m3. Estimate the total electric-field energy stored in the wall of a cell of this size. (Hint: Assume that the cell is spherical, and calculate the volume of the cell wall.) (d) In reality, the cell wall is made up, not of air, but of tissue with a dielectric constant of 5.4. Repeat parts (a) and (b) in this case.

Short Answer

Expert verified

(a) The magnitude of Ein the wall between two layers of charge is 5.6107V/m.

(b) The potential difference between the inside and the outside of the cell is 0.28V.

(c) The total electric field energy stored is 710-15J.

(d) The magnitude of E鈬赌and the potential difference when air is replaced by dielectric constant of 5.4 are and 1.0107V/m0.052Vrespectively.

Step by step solution

01

(a) Determination of magnitude of E→ in the wall between two layers of charge.. 

The cells carrying charges resemble a parallel plate capacitor.

For the medium air, the expression of electric field is,

E0=0=0.50103C/m28.8541012C2/Nm2=5.6107V/m

Thus, magnitude of Ein the wall between two layers of charge is 5.6x107V/m.

02

(b) Determination of the potential difference between the inside and the outside of the cell.

The potential energy expression is given as,

V0=E0d=5.6107V/m5.0109m=0.28V

Thus, the voltage difference is 0.28 V with the outer wall at higher potential.
03

(c) Determination of the total electric field energy.

The energy density expression is,

u=120E02 鈥(颈)

Volume of the cell is equal to volume of a sphere (assumption),

Vcell=43r3

Solve for r,

r=3Vcell413=31010m3413=2.9106m

The volume of the cell wall with thickness t is given as,

VwaII=4R2t=42.9106m25.0109m=5.31019m3

Substitute all the values in equation (i),

u=128.8541012C2/Nm25.6107V/m2=1.39104J/m3

Thus, the total field energy on the cell wall is,

U=uVwal=1.39104J/m35.31019m3=71015J

04

(d) Determination of the magnitude of E→ and the potential difference when air is replaced by dielectric constant of 5.4.

When air is replaced by dielectric the change in electric field is,

E=E0K=5.6107V/m5.4=1.0107V/m

Similarly, for potential difference,

V=V0K=0.28V5.4=0.052V

Thus, the changed electric field and potential difference are 1.0x107V/mand 0.052Vrespectively.

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