Electric Potential

Purple Squirrel and Blue Angel

READ FIRST: THIS WEBSITE (www.myspace.com/webassign or www.webassign.tk) WILL BE OPEN TO THE PUBLIC ONLY FOR THE FIRST TWO WEBASSIGNS. CHANGES TO THE NUMBER OF WEBASSIGNS BEFORE THE WEBSITE CLOSES, IS SUBJECT TO CHANGE AT OUR DISCRETION. THIS WEBSITE WILL BE CLOSED TO PREVENT TEACHERS COMING IN. IF YOU WANT TO HAVE ACCESS TO THE WEBSITE AFTER IT “CLOSES” YOU MUST REQUEST US TO ADD YOU AS FRIENDS. LEGEND IN EQUATIONS (WORK SHOWN): BLUE—COMMON NUMBERS (same for everyone) GREEN—FOUND IN EXPLANATION PURPLE— ANSWER RED— YOUR PERSONAL NUMBERS 1. Suppose that the electric potential outside a living cell is higher than that inside the cell by 0.076 V. How much work is done by the electric force when a sodium ion (charge = +e) moves from the outside to the inside?

The question says the electric potential. The electric potential is the same thing as saying ΔV (measured in volts). If you read the chapter in page 560 it says that: ΔV = -WAB / qo The charge that it talks about is about a positive electron, which is different than a negative electron but they have the same charge: 1.6

×10-19 C

Since you have the change in electric potential, and the charge, you can solve for work. ΔV × qo = W Plug the numbers in: .076 × (1.6×10-19) = 1.216×10-20 C

2. Just as you touch a metal door knob, a spark of electricity (electrons) jumps from your hand to the knob. The electric potential of the knob is 3.0 104V greater than that of your hand. The work done by the electric force on the electrons is 1.5 10-7J. How many electrons jump from your hand to the knob?

In this problem they give you the ΔV and the Work done by the electrons. The only variable missing is the charge. Solve for the charge: ΔV = W/q ΔV × q = W q = W / ΔV Plug the numbers in: (1.5×10-7) × (3.0 × 104) = 5×10-12 C With the charge you can use the charge of a single electron to find the number of electrons that “jumped” to your hand. 5×10-12 C / 1.6×10-19 = 31250000 electrons

3. The anode (positive terminal) of an X-ray tube is at a potential of +105000 V with respect to the cathode (negative terminal). (a) How much work (in joules) is done by the electric force when an electron is accelerated from the cathode to the anode? (b) If the electron is initially at rest, what kinetic energy does the electron have when it arrives at the anode? Question 3. Part a, says: How much work (in joules) is done by the electric force when an electron…… Again, like the previous questions you are going to have to use the ΔV equation. (*Note: ΔV has many different expressions but for these problems we are using the ΔV= W / q) You have the change in electric potential and the charge of an electron, you can use those to plug them in to the equation: ΔV = W / q therefore: ΔV × q = W Plug the numbers in: (105000) × (1.6 × 10-19) = 1.68 × 10-14 Joules In part (b) the question asks for the kinetic energy when the electron arrives at the anode. If you look in the book in page 563, Figure 19.6 you see the TOTAL ENERGY that an object can have. If you take that equation and use the Ef=Eo you will see that almost all of the “terms” in the equation go away and you are left with: 1 /2mvf2 = EPE This is due to the fact that the right side of the equation is the initial energy, since the electron is at rest it doesn’t have any kinetic, since it can’t rotate (going in a straight line) it doesn’t have rotational energy, (we ignore gravity), and no elastic forces are present.

Since EPE (the answer for part a) is the same as kinetic energy the answer for part (b) is the same as part (a) 4. In a television picture tube, electrons strike the screen after being accelerated from rest through a potential difference of 23000 V. The speeds of the electrons are quite large, and for accurate calculations of the speeds, the effects of special relatively must be taken into account. Ignoring such effects, find the electron speed just before the electron strikes the screen.

For question 4, you are going to use the Ef =Eo . The object is only going to have kinetic energy at the end, right before hitting the screen. Therefore the final energy is equal to (1/2mv2). The initial is going to be equal to EPE. We now have: 1 /2mv2 = EPE If you solve to v you get: _____________ v = -\/( EPE × 2) / (m) Plug in the number (EPE): _________________________ -\/ (23000 × 2) / ( 9.109 × 10-31) = 89888380.55 m/s

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