Respuesta :
Answer:
5.3 x 10⁻⁹ C
Explanation:
r = radius of cylindrical shell = 10⁻⁵ m
L = length = 0.32 m
A = area
Area is given as
A = 2πrL
A = 2 (3.14) (10⁻⁵) (0.32)
A = 20.096 x 10⁻⁶ m²
d = separation = 10⁻⁸ m
[tex] k_{appa} [/tex] = dielectric constant = 4
Capacitance is given as
[tex]Q=\frac{k_{appa}\epsilon _{o}A}{d}[/tex] eq-1
V = Potential difference across the membrane = 74 mV = 0.074 Volts
Q = magnitude of charge on each side
Magnitude of charge on each side is given as
Q = CV
using eq-1
[tex]Q=\frac{k_{appa} \epsilon _{o}AV}{d}[/tex]
Inserting the values
[tex]Q=\frac{4 (8.85\times 10^{-12})(20.096\times 10^{-6})(0.074)}{10^{-8}}[/tex]
Q = 5.3 x 10⁻⁹ C
The membrane is parallel and act as a parallel plate capicitor.The magnitude of the charge on each side of the membrane will be 5.3×10⁻⁹C.
What is parallel plate capacitor ?
It is an type capacitor is an in which two metal plates arranged in such away so that they are connected in parallel andhaving some distance between them.
A dielectric medium is must in between these plates help to stop the flow of electric current through it due to its non-conductive nature .
The given data in the problem is;
r is the radius of cell of thin cyliner=10⁻⁵m
L is the length =0.32
d is the thickness =10⁻⁸m
K is the dielectric constant=4
v is the potential difference across the membrane=74 mv=0.074 v
q is the megnitude of charge=?
A is the area of capicitor=2πrl
[tex]\rm A = 2\pi rl\\\\\rm A= 2\times3.14\times10^{-5}\times0.32\\\\\rm A= 20.096\times10^{-6}[/tex]
The given formula for the parallel plate capicitor as ,
[tex]\rm Q=\frac{K\varepsilon _0AEr}{d} \\\\\rm Q=\frac{4\times(8.85\times10^{-12})(20.096\times10^{-6}\times0.074}{10^{-8}}\\\\\rm Q=5.3\TIMES10^{-9}\;C[/tex]
Hence the charge on each side of the membrane will be 5.3×10⁻⁹C.
To learn more about the parallel plate capicitor refer to the link;
https://brainly.com/question/12883102
