# Energy stored in capacitance

1. Apr 12, 2009

### rayhan619

1. The problem statement, all variables and given/known data

The giant axon of a squid is 0.5 mm in diameter, 10 cmlong, and not myelinated. Unmyelinated cell membranes behave as capacitors with 1 microF capacitance per square centimeter of membrane area.
When the axon is charged to the -70 mV resting potential, what is the energy stored in this capacitance?

2. Relevant equations

R = (density* length)/Area

3. The attempt at a solution

I know how to get R and C is given. and V is also given.
but how do i solve for E?

2. Apr 12, 2009

### Redbelly98

Staff Emeritus

1. R is irrelevant here.

2. The capacitance is not 1 microF, in case that is what you are thinking. Read what it says carefully.

3. Your text book should have a discussion of energy stored in a capacitor, along with an equation.

3. Apr 12, 2009

### rayhan619

Energy stored in a capacitor, U = 1/2(Q^2/C) = 1/2(Q/V) = 1/2 CV^2
If we know any two of Q,C, or V we can figure out the energy.
C is 1 microF per square centimeters. we can get the area using A=(3.14)r^2 and get the C from there.
and the V is given -70 mV.
so we can use V and C to get U.
right?

4. Apr 13, 2009

### Redbelly98

Staff Emeritus
Almost. Everything except for A is correct.

The area is a cylinder, not a circle. Use the surface area of a cylinder to get A, and you'll have it.

5. Apr 13, 2009

### rayhan619

so area for cylinder is, A = 2 pi r ( r + h ).
= 2(3.14)(2.5*10^-4){(2.5*10^-4)+0.1} m^2
= 1.57*10^-4 m^2
= 1.57 cm^2
so C = (1*10^-6 F)(1.57 cm^2)
= 1.57*10^-6 F
and V = -70 mV = -70*10^-3 V

Energy, E = 1/2 CV^2 = 1/2 (1.57*10^-6 F)(-70*10^-3 V)^2
= - 4 *10^-9 J

Does that look right?

6. Apr 13, 2009

### Redbelly98

Staff Emeritus
Looks good except that energy is positive.

(-70 mV)^2 = +4900 mV^2 > 0​

7. Apr 13, 2009

### rayhan619

oh ya. thanks