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

Answer:

Part a)

$\Delta V = 72481.2 Volts$

Part b)

$KE = 2.32 \times 10^{-14} J$

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

$\Delta V = E. \Delta x$

$\Delta V = (8.36 \times 10^6)(8.67 \times 10^{-3})$

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

$KE = (2\times 1.6 \times 10^{-19})(72481.2)$

$KE = 2.32 \times 10^{-14} J$