Answer:
[tex]10.99\times 10^6\ V/m[/tex]
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:
[tex]E=-\frac{\Delta V}{\Delta x}[/tex]
So, the magnitude of the electric field is calculated by ignoring the negative sign and thus is given as:
[tex]|E|=\frac{\Delta V}{\Delta x}[/tex]
Plug in the given values and solve for '|E|'. This gives,
[tex]|E|=\frac{0.078\ V}{7.1\times 10^{-9}\ m}\\\\|E|=10.99\times 10^6\ V/m[/tex]
Therefore, the magnitude of the electric field in the membrane is [tex]10.99\times 10^6\ V/m[/tex].
