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 .

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