Question

A bar Membrane walls of living cells have surprisingly large electric fields across them due to separation of ions. What is the voltage across an 8.67-mm-thick membrane if the electric field strength across it is 8.36 MV/m? You may assume a uniform E-field. V helium nucleus has two positive charges and a mass of 6.64 10-27 kg. (a) Calculate its kinetic energy in joules at 7.30% of the speed of light.

Answer 1

Part a)

`\Delta V = 72481.2 Volts`

Part b)

`KE = 2.32 * 10^(-14) J`

Step-by-step explanation:

As we know that the electric field between the membrane is given as

`E = 8.36 MV/m`

here the distance between two membranes is given as

`\Delta x = 8.67 mm`

now we know that potential difference between two membranes is given as

`\Delta V = E. \Delta x`

`\Delta V = (8.36 * 10^6)(8.67 * 10^(-3))`

so we have

`\Delta V = 72481.2 Volts`

Part b)

Kinetic energy of helium nucleus is given as

`KE = Q\Delta V`

so we have

`KE = (2* 1.6 * 10^(-19))(72481.2)`

`KE = 2.32 * 10^(-14) J`