The inner and outer surfaces of a cell membrane carry a negative and positive charge, respectively. Because of these charges, a potential di

Question

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?

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Xavia 4 years 2021-08-22T04:40:46+00:00 1 Answers 12 views 0

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  1. thaiduong
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    2021-08-22T04:42:44+00:00
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    Answer:

    10.99\times 10^6\ V/m

    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:

    E=-\frac{\Delta V}{\Delta x}

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

    |E|=\frac{\Delta V}{\Delta x}

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

    |E|=\frac{0.078\ V}{7.1\times 10^{-9}\ m}\\\\|E|=10.99\times 10^6\ V/m

    Therefore, the magnitude of the electric field in the membrane is 10.99\times 10^6\ V/m.

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