Question

Rolling down the electric hill Distributions of electric charges in a cell play a role in moving ions into and out of a cell. In this situation, the motion of the ion is affected by two forces: the electric force due to the non uniform charge distribution in the cell membrane, and the resistive force (viscosity) due to colliding with the fluid molecules. In order to begin our analysis of this, let's consider a toy model in which the ion is moving in response to electric forces alone Charge in a cell membrane are distributed along the opposite sides of the membrane approximately uniformly. This leads to an (on the average) constant electric field inside the membrane. A simple model that gives this kind of field is two large parallel plates close together. The field between the plates is approximately constant pointing from the negative to the parallel plate. This results in a charge feeling a constant force anywhere between the plate (sort of like flat-earth gravity turned sideways). Outside of the plates the electric fields from the two plates cancel and there is no force A. A sodium ion enters the channel. As it is accelerated, it will pass through the points marked a and b. At which point will the magnitudes of the electric force felt by the ion, the acceleration of the ion, and the electric potential energy of the ion be greater? Check one answer from each column Na Electric force Electric potential energy U, Ub U,= Ub Acceleration Fa Fb Fa F aa ab aa ah Fa F aa ab Not enough info Not enough info Not enough info

Answer 1

Electric - as, force F=&E & and E does not change same Acceleration, as force is same a con acceleration is also [Ma = ab ] Posential energy decrease at it moves towards Negeline plate

Part- (d) electric force F=&E for sodium in &= 1.680079 c F = 16x16 19x 107 a TF = 1.6810-12/N. aueleration a=t mass of sodium ion m= 3.817810-26 kg a (6 8012 3.617 X10-26 a = 4. 1994x1013 m/s2 Ta = 41994 nm/s2 Part-(g) Potential Energy difference Du = 2 (V2_W) su= 2 (AV) DV= - Ed AU=&Ed DU = -1.6X10 13 x 107x7X10-9 ou =-1112 X10-20 5 Part-(f) gain in OKE= - (AU) TOKE = 1.12710-20/J