Q19.2-18PE - College Physics (Urone)

Question

The voltage across a membrane forming a cell wall is \(80.0{\rm{ }}mV\) and the membrane is \(9.00{\rm{ }}nm\) thick. What is the electric field strength? (The value is surprisingly large, but correct. Membranes are discussed in Capacitors and Dielectrics and Nerve Conduction—Electrocardiograms.) You may assume a uniform electric field.

Step-by-Step Solution

Answer

Value of the electric field strength is \(8.89 \times {10^6}\;V/m\).

Step 1: Information Provided

• Voltage across a membrane is: \(80.0{\rm{ }}mV\)
• Thickness of the membrane is: \(9.00{\rm{ }}nm\)

Step 2: Relation between potential and electric field and calculation for potential and thickness of membrane.

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

 

Now, calculate voltage across the membrane:

\(\begin{array}{c}\Delta V = (80.0mV)\left( {\frac{{1\;V}}{{1000\;V}}} \right)\\\Delta V = 80.0 \times {10^{ - 3}}\;V\end{array}\)

Now calculation for the thickness of the membrane:

\(\begin{array}{c}d = (9.00\;nm)\left( {\frac{{1\;m}}{{{{10}^9}\;nm}}} \right)\\d = 9.00 \times {10^{ - 9}}\;m\end{array}\)

Step 3: Calculation for the electric strength in the membrane.

The electric field strength in the membrane is found by solving Equation for \(E\):

\(E = \frac{{\Delta V}}{d}\)

Entering the values for \(\Delta V\) and \(d\), we obtain:

\(\begin{array}{c}E{\rm{ }} = \frac{{80.0 \times {{10}^{ - 3}}\;V}}{{9.00 \times {{10}^{ - 9}}\;m}}\\E = 8.89 \times {10^6}\;V/m\end{array}\)

 

Therefore, the value is obtained as \(8.89 \times {10^6}\;V/m\).