Can anyone help with the following 2 questions over electrical energy and capacitance? 1) To recharge a 12-V battery, a battery charger must move 3.6x10^5C of charge from the neg. to pos. terminal. How much work is done by the charger in joules? 2) Calculate the speed of a proton that is accelerated from rest through a potential difference of 120V. What is the speed of an e-? I know this is fairly simple to some, but I'm having major trouble! Any help (or strategy) would be appreciated.

Electrical work can easily be found by taking the amount of charge and multiplying it by the potential difference (in this case 12V). You can think of this like moving a brick from the ground to the top of your desk. The amount of work required is equal to the weight of the brick (equivalent to amount of charge in the previous problem) times the height of the desk (equivalent to the potential difference). The speed of a charged particle in an electrical field can be found simply by determing the amount of kinetic energy it has once it has accelerated through the field. An electron or proton accelerated through 120 V will have a kinetic energy of 120 eV where 1 eV = 1.6e-19 J. Kinetic energy is just equal to 1/2 * m * v^2. You can find the mass of the proton/electron and solve for velocity. If the energy is in Joules and mass in kg, the velocity will be in m/s.

I know the second one. Someone (not some one--Kutastha is even more anal than I am about spelling and grammar) can do the other one. W=delta K = q times abs. val delta v 1/2 mvsquared=120V times e so v = square root of around 3.8 times 10 to the -17 J divided by 1.67 times 10 to the -27 kg = (your answer) 152 km/s Oops you want an e-=so divide instead by 9.11 times 10 to the -31 and you get 6.49 times 10 to the 6 m/s

ok, so that leaves me to do the first one. I believe the two just have to be multiplied to get the answer. Think about it this way, they have given you VOLTS and COULOMBS which is AMPS/Second, right? So, if Power = Work/Time which would be in Joules/time units, and we already have "per time" all you have to do is multiply the two numbers to get Joules. Right? P = Work/Time therefore, Coulombs * Volts/Time = work. God I hope that's right. Tweetie

You guys are awesome! I really appreciate your help and time (and explanations)! <img src="graemlins/clappy.gif" border="0" alt="[Clappy]" />

I came to this thread with the intention of helping you do the physics problems...but quickly realized that I don't remember how to do any of those problems. I have a vague recollection..but that's about it. Physics was always my weakest subject. Here's the kicker...I got a 12 on Physics section. Keep up the hard work...it will pay off.