Answer:
A) 1×10^6N/C
B) 2×10^-13N
C) 1×10^-2v
Explanation:
For parallel plates,the electric field E is given by:
E=s/eo
Where s= surface charge density
E = 10^-5/8.85×10^-12
E= 1.13×10^6 approximately 1×10^6NC^-2
B) Na has a charge of 1.6×10^-19
F= q×E= (1.13×10^6) × (1.6×10^-19)
F= 1.8×10^-13 approximately 2×10^-13N
C) Potential difference ,V= E×d
d=10nm=10^-8
V= 1.13×10^6 ×10^-8
V = 1.13×10^-2 approximately 1x10^-2v
We have that for given the wall separated by a distance of approximately 10nm. and charge density of 10?5C/m2
a) Electric field
E=1.13*10^6N/c
b) Force
F=1.808*10{-13}N
c) the Voltage
v=1.13×1014v
From the question we are told
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act like a parallel plate capacitor, each with a charge density of 10?5C/m2, and the outer wall is positively charged. Although unrealistic, assume that the space between cell walls is filled with air.
a
Generally the equation for the electric field is mathematically given as
[tex]E=\frac{d}{e}\\\\Therefore\\\\E=\frac{10^{-5}}{8.85*10^{-12}}\\\\E=1.13*10^6N/c[/tex]
b
Generally the equation for the Force is mathematically given as
[tex]F=qE\\\\Where\\\\Charge is K+\\\\F=1.6*10^{-19}*1.13*10^6N\\\\F=1.808*10{-13}N[/tex]
c)
Generally the equation for the Voltage is mathematically given as
[tex]v=\frac{E}{d}[/tex]
[tex]v=\frac{1.13*10^6}{10^10^{-9}}[/tex]}
v=1.13×1014v
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