The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 4.85×10−9 m2 , a plate separation of 8.2×10−9 m, and a dielectric with a dielectric constant of 4.6.What is the energy stored in such a cell membrane if the potential difference across it is 7.55×10−2 V?

To find the energy stored in the parallel-plate capacitor, we can use the formula for the energy stored in a capacitor:

\[
U = \frac{1}{2} C V^2
\]

where:
- \( U \) is the energy stored,
- \( C \) is the capacitance of the capacitor,
- \( V \) is the potential difference across the capacitor.

### Step 1: Calculate the Capacitance

The capacitance \( C \) of a parallel-plate capacitor is given by:

\[
C = \frac{\varepsilon_0 \varepsilon_r A}{d}
\]

where:
- \( \varepsilon_0 \) is the permittivity of free space, \( \varepsilon_0 = 8.854 \times 10^{-12} \, \text{F/m} \),
- \( \varepsilon_r \) is the dielectric constant,
- \( A \) is the area of the plates,
- \( d \) is the separation between the plates.

Plugging in the given values:

- \( \varepsilon_r = 4.6 \)
- \( A = 4.85 \times 10^{-9} \, \text{m}^2 \)
- \( d = 8.2 \times 10^{-9} \, \text{m} \)

\[
C = \frac{(8.854 \times 10^{-12} \, \text{F/m}) \times 4.6 \times 4.85 \times 10^{-9} \, \text{m}^2}{8.2 \times 10^{-9} \, \text{m}}
\]

Calculating \( C \):

\[
C = \frac{(8.854 \times 10^{-12}) \times 4.6 \times 4.85}{8.2} \times 10^{-12} \, \text{F}
\]

### Step 2: Calculate the Energy Stored

Now, using the formula for energy stored:

\[
U = \frac{1}{2} C V^2
\]

where \( V = 7.55 \times 10^{-2} \, \text{V} \).

Let’s calculate the energy stored.

I'll do the calculations for both capacitance and energy stored.### Calculated Results:
1. **Capacitance (\( C \))**: \( 2.41 \times 10^{-11} \, \text{F} \)
2. **Energy Stored (\( U \))**: \( 6.87 \times 10^{-14} \, \text{J} \)

The energy stored in the cell membrane, given the provided parameters, is approximately \( 6.87 \times 10^{-14} \, \text{J} \).

Would you like to know more details or have any questions?

### Related Questions:
1. How does the energy stored change if the plate separation is halved?
2. What effect does doubling the dielectric constant have on the energy stored?
3. How would the energy stored change if the potential difference is doubled?
4. What is the physical significance of the dielectric constant in a capacitor?
5. How can the concept of a parallel-plate capacitor be applied in practical biological systems?

**Tip:** In a capacitor, increasing the dielectric constant \( \varepsilon_r \) increases the capacitance, allowing more energy to be stored for the same voltage.

The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 4.85×10−9 m², a plate separation of 8.2×10−9 m, and a dielectric with a dielectric constant of 4.6. What is the energy stored in such a cell membrane if the potential difference across it is 7.55×10−2 V?

Learn how to calculate the energy stored in a cell membrane modeled as a parallel-plate capacitor with a dielectric constant of 4.6. This problem includes the calculation of capacitance and energy stored using key physics concepts and formulas. Suitable for undergraduate physics students.

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