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Readme [11.4K]
3 years ago
5

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 0.063 V exists across the membrane. The thickness of the membrane is 9.9 x 10-9 m. What is the magnitude of the electric field in the membrane
Physics
1 answer:
sertanlavr [38]3 years ago
7 0

Answer:

Electric field, E = 6.36\times 10^{6}\ V/m

Explanation:

Given that,

The potential difference across the membrane, \Delta V=0.063\ V

The thickness of the membrane is, \Delta x=9.9\times 10^{-9}\ m

We need to find the magnitude of the electric field in the membrane. We know that the relation between electric field and electric potential is given by :

E=\dfrac{-\Delta V}{\Delta x}\\\\E=\dfrac{0.063\ V}{9.9\times 10^{-9}\ m}\\\\E=6.36\times 10^{6}\ V/m

So, the magnitude of the electric field in the membrane is 6.36\times 10^{6}\ V/m.

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Answer:

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Explanation:

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F_N=\sqrt{F^{2}+F^{2}}\\\\F_N=\sqrt{2F^{2}}\\\\F_N=\sqrt{2}F

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A uranium ion and an iron ion are separated by a distance of =61.10 nm. The uranium atom is singly ionized; the iron atom is dou
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Answer:

Explanation:

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charge on doubly ionised iron atom = charge of 2 electron

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Let the required distance from uranium ion be d .

force on electron at distance d from uranium ion

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For equilibrium ,

9 x 10⁹ x 1.6 x 10⁻¹⁹ / r² = 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

2 d² = (61.10 x 10⁻⁹ - r )²

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2.414 r = 61.10 x 10⁻⁹

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Answer:

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For first wire length is l₁ , area is A₁ resistance is R₁, for second resistance is R₂ , length is l₂ and area is A₂

Given , l₁ = 2l₂ , A₁ = 4A₂ , area is proportional to square of thickness.

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