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yaroslaw [1]
3 years ago
5

The voltage across a membrane forming a cell wall is 72.7 mV and the membrane is 9.22 nm thick. What is the magnitude of the ele

ctric field strength? (The value is surprisingly large, but correct.) You may assume a uniform E-field.
Physics
1 answer:
Blizzard [7]3 years ago
8 0

Answer:

The  magnitude of the  electric field intensity is  E =  7.89  *10^{6} \ V/m

Explanation:

From the question we are told that

    The  voltage is  \epsilon     =  72.7 \ mV  =  72.7 *10^{-3}  V

    The  thickness of the membrane is  t =  9.22 \ nm  =  9.22 *10^{-9} \ m

     

Generally the electric field intensity is mathematically represented as

                E =  \frac{\epsilon }{t}

 substituting values

                E =  \frac{72.7 *10^{-3} }{9.22 *10^{-9}}

                E =  7.89  *10^{6} \ V/m

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Answer:

q = 2.066* 10⁻¹³ C.

n = 1,291,250 electrons.

Explanation:

1)

  • If the gravitational attraction is equal to their electrical repulsion, we can write the following equation:

       F_{g} = F_{c} (1)

  • where Fg is the gravitational attraction, that can be written as follows        according Newton's Universal Law of Gravitation:

       F_{g} = G*\frac{m_{1}*m_{2}}{r_{12}^{2}} (2)

  • Fc, due to it is the electrical repulsion between both charged spheres, must obey Coulomb's Law (assuming  we can treat both spheres as point charges), as follows:

       F_{c} = k*\frac{q_{1}*q_{2}}{r_{12}^{2}} (3)

  • since m₁ = m₂ = 0.0024 kg, and  r₁₂ = 0.1m, G and k universal constants, and q₁ = q₂ = Q, we can replace the values in (2) and (3), so we can rewrite (1) as follows:

       G*\frac{(0.0024kg)^{2}}{r_{12}^{2}} = k*\frac{Q^{2}}{r_{12}^{2}} (4)

  • Since obviously the distance is the same on both sides, we can cancel them out, and solve (4) for Q² first, as follows:

       Q^{2} = \frac{6.67e-11*(0.0024kg)^{2}}{9e9Nm2/C2} = 4.27*e-26 C2 (5)

  • Since both charges are the same, the charge on each sphere is just the square root of (5):
  • Q = 2.066* 10⁻¹³ C.

2)

  • Assuming that both spheres were electrically neutral before being charged, the negative charge removed must be equal to the positive charge on the spheres.
  • Now, since each electron carries an elementary charge equal to -1.6*10⁻¹⁹ C, in order to get the number of electrons removed from each sphere, we need to divide the charge removed from each sphere (the outcome of part 1) with negative sign) by the elementary charge, as follows:
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4 0
3 years ago
The equipotential surfaces surrounding a point charge are concentric spheres with the charge at the center. If the electric pote
tester [92]

Answer:

1.06 m

Explanation:

Since the charge is at the centre of two concentric spheres, we use the formula for electric potential due to a point charge. V = kq/r. Let r₁ be the radius of the sphere with potential, V₁ = 200 V and  r₂ be the radius of the sphere with potential, V₂ = 82.0 V. From V = kq/r, r = kq/V. So that r₁ = kq/V₁ and r₂ = kq/V₂. The magnitude of the difference r₁ - r₂ is the distance between the two surfaces. q the charge equals 1.63 × 10⁻⁸ C

r₂ - r₁ = kq/V₂ - kq/V₁ = kq(1/V₂ - 1/V₁) = 1.63 × 10⁻⁸ × 9 × 10⁹ (1/82 -1/200) m =  1.63 × 10⁻⁸ × 9 × 10⁹ (0.0122 - 0.005) = 1.63 × 10⁻⁸ × 9 × 10⁹(0.0072) m = 1.06 m

The distance between them is 1.06 m

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Answer:

Ok

Explanation:

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8 0
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Strike441 [17]

Answer:

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