The inner and outer surfaces of a cell membrane carry a negative and positive charge respectively. Because of these charges, a potential difference of about 70*10-3 V exists across the membrane. The thickness of the membrane is 8.0*10-9 m. Assumption the membrane has a dielectric constant close to 1.0, the magnitude of the electric field inside the membrane would be _________ * 106 V/m

To determine the magnitude of the electric field inside the membrane, we can use the formula:

E = V / d

Where:
E is the magnitude of the electric field,
V is the potential difference across the membrane, and
d is the thickness of the membrane.

Given:
V = 70 * 10^(-3) V (potential difference)
d = 8.0 * 10^(-9) m (thickness of the membrane)

Plugging in the given values into the equation, we get:

E = (70 * 10^(-3) V) / (8.0 * 10^(-9) m)

To simplify the calculation, we can divide the numerator and denominator separately:

E = 70 / 8.0 * (10^(-3) / 10^(-9)) V/m

Simplifying the exponents:

E = 70 / 8.0 * 10^6 V/m

Therefore, the magnitude of the electric field inside the membrane is 8.75 * 10^6 V/m.