Question

When a nerve cell depolarizes, charge is transferred across the cell membrane, changing the potential difference. For a typical nerve cell, 9.0 pC of charge flows in a time of 0.50ms . What is the average current? 

Answer 1

Concepts and reason

The main concepts required to solve this problem are the current, charge, and time.

Initially, write the equation for the current passing through the conductor per unit time. Use this equation and calculate the average current through the cell membrane.

Fundamentals

The equation for the current passing through the conducting wire per unit time is,

$I = \frac{Q}{t}$

Here, Q is the amount of charge, and t is the time taken by the charge to pass through the conductor between the two potentials.

Electric current can be defined as the amount of charge that passing through the conducting wire between the potential differences per unit time. The electric current can be denoted by the letter I. The current can be measured with the unit of Ampere, that is A.

The expression for the current that passing through the conductor is,

$I = \frac{Q}{t}$

Here, Q is the amount of charge, and t is the time taken by the charge to pass through the conductor between the two potentials.

The equation for the current passing through the cell membrane that found in the above step1 is,

$I = \frac{Q}{t}$

Here, Q is the charge flowing through the membrane, and t is the time.

Substitute $9.0{\rm{ pC}}$ for Q, and $0.50{\rm{ ms}}$ for t in above equation.

$\begin{array}{c}\\I = \frac{{9.0{\rm{ pC}}\left( {\frac{{{{10}^{ - 12}}{\rm{ C}}}}{{1{\rm{ pC}}}}} \right)}}{{0.50{\rm{ ms}}\left( {\frac{{{{10}^{ - 3}}{\rm{ s}}}}{{1{\rm{ ms}}}}} \right)}}\\\\ = 18 \times {10^{ - 9}}{\rm{ C}}\left( {\frac{{1{\rm{ nA}}}}{{{{10}^{ - 9}}{\rm{ A}}}}} \right)\\\\ = 18{\rm{ nA}}\\\end{array}$

Ans:

The average current passing through the membrane is $18{\rm{ nA}}$ .