**Problem Statement:**The membrane that surrounds a certain type of living cell has a surface area of \(3.8 \times 10^{-9} \, \text{m}^2\) and a thickness of \(1.4 \times 10^{-8} \, \text{m}\). Assume that the membrane behaves like a parallel plate capacitor and has a dielectric constant of 4.6.(a) The potential on the outer surface of the membrane is +46.4 mV greater than that on the inside surface. How much charge resides on the outer surface?(b) If the charge in part (a) is due to \( \text{K}^+ \) ions (charge \( +e \)), how many such ions are present on the outer surface?**Answer:**(a) Number: \(0.00000000005127\)Units: C(b) Number: \(3204458571\)Units: No units

Let the membrane that surrounds a certain type of living cell has a surface area,(A) in metre. So,…

Answered: The membrane that surrounds a certain…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Consider a parallel-plate capacitor with a plate area of A = 8.50 cm². The separation between the plates is di = 3.00 mm (the space between the plates is filled with air). The plates of the capacitor are charged by a 6.00 V battery, i.e., the potential difference between the plates is V₂ = 6.00 V. The plates are then disconnected from the battery and pulled apart (without discharge) to a sepa- ration of df = 8.00 mm. In the following, neglecting any fringing effects. (a) Will the new potential difference between the plates be larger, smaller, or the same compared to the initial potential difference of Vi = 6.00 V? Explain. (Hint: Note that the charge will not change when the plates are pulled apart. Why is that?) (b) Find the potential difference Vf between the plates after the plates have been pulled to their new, larger separation df. (c) Find the electrostatic energy stored in the capacitor before and after the plates are pulled apart. (d) To separate the plates, you will need to do…
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.70 mm. If a 17.8-V potential difference is applied to these plates, calculate the following. (a) the electric field between the plates; magnitude and direction (b) the capacitance (c) the charge on each plate
The potential in a region between x = 0 and x = 6.00 m is V = a + bx, where a = 15.2 V and b = -3.50 V/m. a) Determine the potential at x= 0, x= 3.00 m, and x= 6.00 m. b) Determine the magnitude (Vm) and direction of the electric field at x= 0, x= 3.00 m, and x= 6.00 m.
Problem 18: The electric potential in a certain region is given by the equation V(x,y,z) = 3αx2y3 - 2γx2y4z2 where the potential is in volts when the positions are given in meters. Part (d) The constants in this equation are α = 5.2 V/m5 and γ = 1.39 V/m8. Using this information, calculate the magnitude of the electric field at the point (x1,y1,z1) = (-5.0, 2.0, 1.5) m in units of newtons per coulomb.
A parallel plate capacitor is composed of two square plates, 8.00 cm on a side, separated by a layer of dielectric of thickness 0.600 mm. The dielectric constant of the dielectric material is κ. a)If the capacitance of the capacitor is C = 2.83 × 10−10 F, then what is κ? (b) If the charge held on the capacitor’s plates is Q = 5.66 nC, then what is the potential difference between the plates? (c) What is the energy stored in the capacitor?
TWO CONDUCTING SPHERES, ONE WITH A RADIUS OF 5 cm AND THE OTHER WITH A RADIUS OF 11 cm, EACH OF THEM CONTAIN EQUAL CHARGES WHOSE VALUE IS Q=28 nC AND THEY ARE VERY FAR AWAY FROM EACH OTHER. IF THE SPHERES ARE LATER CONNECTED BY MEANS OF A CONDUCTING WIRE, FIND a) THE FINAL LOAD ON EACH SPHEREb) THE ELECTRICAL POTENTIAL OF EACH OF THEM
A potential difference exists between the inner and outer surfaces of the membrane of a cell. The inner surface is negative relative to the outer surface. If 1.25 x 10^-20 J of work is required to eject a positive sodium ion (Na+) from the interior of the cell, what is the magnitude of the potential difference between the inner and outer surfaces of the cell?
What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.16 m whose potential is 300 V (with V = 0 at infinity)? (a) Number Units (b) Number Units
An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.70 mm. A 23.0-V potential difference is applied to these plates. (a) Calculate the electric field between the plates. kV/m(b) Calculate the surface charge density. (c) Calculate the capacitance. (d) Calculate the charge on each plate.
A uniform electric field is oriented in the -x direction. The magnitude of the electric field is 6500 N/C. (a) How will the equipotential surfaces associated with this electric field be oriented? O parallel to the xy plane O parallel to the xz plane O parallel to the yz plane (b) Consider two of the equipotential surfaces; one with a potential of 11 V and the other with a potential of 12 V. What is the separation between these two surfaces? mm
11 of 35 > A potential difference exists between the inner and outer surfaces of the membrane of a cell. The inner surface is negative relative to the outer surface. If 1.25 x 10-20 J of work is required to eject a positive sodium ion (Na+) from the interior of the cell, what is the magnitude of the potential difference between the inner and outer surfaces of the cell? magnitude of the potential difference:

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