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 0.078 V exists across the membrane. The thickness of the membrane is 7.1 x 10-9 m. What is the magnitude of the electric field in the membrane?

Explanation:

Given:

Potential difference across the membrane (ΔV) = 0.078 V

Thickness of the membrane (Δx) = 7.1 × 10⁻⁹ m

Magnitude of electric field (|E|) = ?

We know that, the electric field due to a potential difference (ΔV) across a distance of Δx is given as:

So, the magnitude of the electric field is calculated by ignoring the negative sign and thus is given as:

Plug in the given values and solve for '|E|'. This gives,

Therefore, the magnitude of the electric field in the membrane is .

answer would be d.) i hope this

explanation:

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .