Q9.3-19E - College Physics (Urone)

Question

Membrane walls of living cells have surprisingly large electric fields across them due to separation of ions. (Membranes are discussed in some detail in Nerve Conduction—Electrocardiograms.) What is the voltage across an \(8.00{\rm{ }}nm\)–thick membrane if the electric field strength across it is \(5.50{\rm{ }}MV/m\)? You may assume a uniform electric field.

Step-by-Step Solution

Answer

Value of the voltage is \(4.40 \times {10^{ - 2}}\;V\).

Step 1: Relation between potential & electric field and Calculation of thickness membrane & electric field across membrane.

The potential difference between two points separated by a distance \(d\) in a homogeneous electric field of magnitude \(E\) is \(\Delta V = Ed{\rm{ }}......(1)\).

Step 2: The given data

Now calculation for the thickness of the membrane:

\(\begin{array}{c}d = (8.00\;nm)\left( {\frac{{1\;m}}{{{{10}^9}\;nm}}} \right)\\d = 8.00 \times {10^{ - 9}}\;m\end{array}\)

Now calculation for the electric field across the membrane is:

\(\begin{array}{c}E = (5.50{\rm{ }}MV/m)\left( {\frac{{{{10}^6}\;V}}{{1{\rm{ }}MV}}} \right)\\E = 5.50 \times {10^6}\;V/m\end{array}\)

Step 3: Calculation for the voltage across membrane.

Now the voltage across the membrane is found by using equation below

\(\begin{array}{c}\Delta V = \left( {5.50 \times {{10}^6}\;V/m} \right)\left( {8.00 \times {{10}^{ - 9}}\;m} \right)\\\Delta V = 4.40 \times {10^{ - 2}}\;V\end{array}\)

 

Therefore, the value of the potential across the membrane is \(4.40 \times {10^{ - 2}}\;V\).