A proton moves parallel to an electric field between points A and B. The change in the proton's electric potential energy is EPEB – EPEA. If the proton is removed from point B and replaced with an electron, how does the change in electric potential energy of the electron compare to the change in electric potential energy of the proton as the electron moves from point B to point A? The change in electric potential energy for the electron is EPEA – EPEB.

The change in electric potential energy for the electron is EPEB – EPEA.

The change in electric potential energy for the electron is EPEB + EPEA.

The change in electric potential energy for the electron is zero joules.

More information is needed concerning the charges involved.

The inside surface of a cylindrical-shaped cave of inner diameter 1.0 m is continuously covered with a very thin layer of water. the cave is very long and it is open on both ends. the water on the cave surface is at a constant temperature of 15.5 °c. the cave is constantly exposed to wind such that 15.5 °c air flows through the cave at 4.5 m/s. the kinematic viscosity of the air is 14.66 x 10-6 m2/s and the molecular diffusion coefficient of water vapor in the air is 0.239 x 10-4 m2/s. because the cave diameter is so large, the flow of wind down the length of the cave, in the x direction, can be treated like it is external flow and the cave surface can be approximated as flat where appropriate. calculate the x value, in a) the transition to turbulent flow occurs at rex meters, where the air flow transitions from laminar to turbulent along the inside surface of the cave b) calculate the x value, in meters, where the bulk steady state concentration of water vapor in the air flowing in the cave is 10% of the saturation concentration. assume the air at the surface of the water layer is 100% saturated with water vapor. assume the wind entering the cave contained no moisture before it entered the cave. take into account the transition from laminar to turbulent flow when solving part b