Question

The membrane that surrounds a certain type of living cell has a surface area of 4.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 4.2. (a) The potential on the outer surface of the membrane is 85.7 mV greater than that on the inside surface. How much charge resides on the outer surface

Answer 1

The charge resides on the outer surface =
`1.245 * 10^(-12)` C

Step-by-step explanation:

Surface area of cell
`(A) = 4.3* 10^(-9) m^(2)`

Separation between two plate
`(d) = 1.1 * 10^(-8) m`

Dielectric constant
`(k) = 4.2`

Potential difference
`(\Delta V) = 85.7 * 10^(-3) V`

The capacitance of parallel plate capacitor in free space is given by,

`C = (\epsilon_(o) A )/(d)`

Where
`\epsilon_(o) =` permittivity of free space =
`8.85 * 10^(-12)`

The Capacitance of capacitor is increase by
`k` times when it placed in dielectric medium.

`C_(dielectric) = (k \epsilon_(o) A )/(d)`

And we know that,
`C = (Q)/( \Delta V)`

So charge on the outer surface is given by,

`Q = (k \epsilon A \Delta V )/(d)`

`Q = (4.2 * 4.3 * 10^(-9) * 8.85 * 10^(-12) * 85.7* 10^(-3) )/(1.1 * 10^(-8) )`

`Q = 1.245 * 10^(-12)`