Answer:
Energy stored in cell membrane is 5.83 x 10⁻¹⁴ J.
Explanation:
Given :
Area of parallel plate capacitor,A = 5.41 x 10⁻⁹ m²
Distance between the plates,d = 8.5 x 10⁻⁹ m
Dielectric constant, k = 4.5
Capacitance of a parallel plate capacitor is given by the relation :
C = (kε₀A)/d
Here ε₀ is vacuum permittivity and its value is 8.85 x 10⁻¹² F/m.
Substitute the suitable values in the above equation.
[tex]C = \frac{4.5\times8.85\times10^{-12}\times5.41\times10^{-19} }{8.5\times10^{-19}}[/tex]
C = 2.53 x 10⁻¹¹ F
Potential difference between the plate, V = 0.0679 V
Energy stored in a capacitor is given by the relation :
[tex]E = \frac{1}{2}CV^{2}[/tex]
Substitute the values of C and V in the above equation.
[tex]E = \frac{1}{2}\times2.53\times10^{-11}\times (0.0679)^{2}[/tex]
E = 5.83 x 10⁻¹⁴ J
