Giancoli Physics (5th ed) Chapter 17
- 1 Problems
- 1.1 1. How much work is needed to move a -8.6C charge from ground to a point whose potential is +75V?
- 1.2 2.
- 1.3 3.How much kinetic energy will an electron gain (in joules and eV) if it falls through a potential difference of 21,000V in a TV picture tube?
- 1.4 4.
- 1.5 5. How strong is the electric field between two parallel plates 5.2mm apart if the potential difference between them is 220V?
- 1.6 6.
- 1.7 7. What potential difference is needed to give a helium nucleus (Q=2e) 65.0keV of KE?
- 1.8 8
- 1.9 9.
- 1.10 10.
- 1.11 11.What is the speed of a proton whose kinetic energy is 28.0 MeV?
- 1.12 12.
- 1.13 13.
- 1.14 14.
- 1.15 15. A +30 micro coulomb charge is placed 32 cm from an identical +30 micro coulomb charge. How much work would be required to move a +0.50 micro coulomb charge from a point midway between them to t point 10 cm closer to either of the two charges?
- 1.16 16.
- 1.17 17.
- 1.18 18.
- 1.19 19.Consider point a which is 70 cm north of a -3.8mC point charge, and point b which is 80 cm west of the charge (Fig. 17-23 in Giancoli). Determine (a) Vba = Vb - Va and (b) Eb - Ea (magnitude and direction).
- 1.20 20.
- 1.21 21. Two identical +7.5C point charges are initially spaced 5.5 cm from each other. If they are released at the same instant from rest, how fast will they be moving when they are very far away from each others? Assume they have identical masses of 1.0 mg.
- 1.22 22.
- 1.23 23.
- 1.24 24.
- 1.25 25.
- 1.26 26.
- 1.27 27.
- 1.28 28.
- 1.29 29.
- 1.30 30.
- 1.31 31.
- 1.32 32.
- 1.33 33.
- 1.34 34.
- 1.35 35.
- 1.36 36.
- 1.37 37.
- 1.38 38.
- 1.39 39.
- 1.40 40.
1. How much work is needed to move a -8.6C charge from ground to a point whose potential is +75V?
(-6.5x10-4J (done by the field)
(-2.40x10-17J (done by the field))
3.How much kinetic energy will an electron gain (in joules and eV) if it falls through a potential difference of 21,000V in a TV picture tube?
There are two ways to do this problem. One will give you the answer in eV, and one will give you the answer in joules. Neither are difficult. The first way is the more complicated, and uses the formula from the previous question, W=qV. V=21000, the potential difference, and q is the charge of one electron, or 1.602E-19C. Therefore, the work (which will all be converted to kinetic energy) is 21000*1.602e-19=3.4E-15J. To get the answer in eV is much easier. We know that it is one electron being accelerated through 21000V. Therefore, the total number of electron Volts will be 21000, or 21keV.
5. How strong is the electric field between two parallel plates 5.2mm apart if the potential difference between them is 220V?
Well, you should know that there are two different units for electric field strength: either N/C, or V/m. These are both in the data packet, as E=F/q and E=-DV/Dx. E=-DV/Dx is obviously the one to use here.
The sign of the answer only shows the direction of the field.
7. What potential difference is needed to give a helium nucleus (Q=2e) 65.0keV of KE?
We know that a single electron, moved through V volts, will gain V electron volts of energy. Similarly, 2 electrons (or an object with the same charge as two electrons, such as a helium nucleus) will gain 2V electron volts of energy when moved through V volts. From this, we can see that an object with a charge of two electrons (or protons) that has 65000eV of energy would have to only be accelerated through half the voltage a single electron would have been. This means that the answer is 65000/2, or 32500V. To do the math: 65000eV/2e=32500V. (-32.5kV)
(Va-Vb = 269 V)
11.What is the speed of a proton whose kinetic energy is 28.0 MeV?
I think that the easiest way to do this would be to use the equation W = .5 * m * v2. First, we must convert 28.0MeV to joules. This is a simple matter. 28.0E6eV * 1.602E-19C/e
So, the proton has 4.49E-12 joules of kinetic energy. Since we know the mass of the proton (1.67E-27kg, for those of you too lazy to actually look at your data packet), it is another simple matter to solve for the velocity.
W = .5 * m * v2
4.49E-12 = .5 * 1.67E-27 * v2
4.49E-12 / .5 / 1.67E-27 = v2
v2 = 5.38E15
SO, v = 7.33E7ms-1
15. A +30 micro coulomb charge is placed 32 cm from an identical +30 micro coulomb charge. How much work would be required to move a +0.50 micro coulomb charge from a point midway between them to t point 10 cm closer to either of the two charges?
All we need to do is multiply the test charge by the change in voltage. V = W/q, so the work, W = Vq where V is the change in voltage, and q is the charge moved.
We now have two locations, one directly between the two +30 micro coulomb charges, that is 16 cm from each of them, since they are 32 cm apart, and a second location that is 6 cm from one of the charges, and 26 cm from the other. See the diagram below:
So to solve these problems you just do these three steps:
- Find the voltage before and after the move
- Find the difference in voltage (subtract)
- Use W = Vq to calculate the work done
So the voltage before is simply V = kq/r + kq/r (yay no vectors)
Vbefore = 8.99e9(30e-6)/.16 + 8.99e9(30e-6)/.16 = 3371250 V
and after it is: (note the distances are the only things that change)
Vafter = 8.99e9(30e-6)/.06 + 8.99e9(30e-6)/.26 = 5532307.692 V
The change is that we went up in voltage from 3371250 V to 5532307.692 V, a change of 2161057.692 volts.
Since we are moving a charge of 0.50 micro coulombs through this difference, the work is simply W = Vq, W = (2161057.692 V)(0.50e-6 C) = 1.08 J
19.Consider point a which is 70 cm north of a -3.8mC point charge, and point b which is 80 cm west of the charge (Fig. 17-23 in Giancoli). Determine (a) Vba = Vb - Va and (b) Eb - Ea (magnitude and direction).
(a) Ok, first we need to find Vb and Va using V=kq/r Vb=8.99E9*-3.8E-6/0.8=-42703V Va=8.99E9*-3.8E-6/0.7=-48803V So, Vba=-42703--48803=6100V
(b) The electric field is a bit more complicated. We will use a vector at each point to represent the electric field strength there and calculate the field strength using E=kq/r2. At point b, E=8.99E9*-3.8E-6/0.82=-53378N/C. Since a positive charge placed at point b will travel to the right (it is to the left of a negative charge), we know that the vector will only have an x-component, and it will be positive. So, the field at that point is represented by 53378x+0y. The point b is similar. E=8.99E9*-3.8E-6/0.72=-69718N/C. The vector at that point will only have a negative y-component, so the vector will be 0x+-69718y.
Moving to the next step, Eb-Ea = (53378-0)x+(0--69718)y=53378x+69718y, in vector component form. (Isn't this great? It's just a chapter 2 problem in disguise!!) The question asks for the answer in angle-magnitude form, though. So, we need to convert. The magnitude is easy. It's just the Pythagorean theorem. 533782+697182=c2, so c=87806N/C. As for the direction, tan-1(69718/53378)=53º from vertical, up and to the right. So in conclusion, the answer is 8.8E4N/C, 53° E of N.
b.(53° N of E)
21. Two identical +7.5C point charges are initially spaced 5.5 cm from each other. If they are released at the same instant from rest, how fast will they be moving when they are very far away from each others? Assume they have identical masses of 1.0 mg.
To calculate this, we will use electrostatic potential energy. Now, we know that Epot=Ekin, because energy is conserved, and the potential energy of the system will equal the amount of work required to put the charges in their current state. We also know that V=kq/r and V=W/q.
V= = = 1225909.09 V
W=Vq = (1225909.09 V)(7.5x10-6 C)
So the total energy of the system is 9.194 joules. Since both charges have the same mass, and momentum is conserved, they will have equal and opposite velocities. To find these, we use good old 1/2*m*v2. Also, because they have the same mass, they will end up with the same amount of energy, 9.194/2, or 4.597J.
yeeeeeaaaaaa... about that...