what potential difference is needed to give a helium nucleus 65 kev of kinetic energy
IB Problem Fix 17: | 1 | three | v | vii | 11 | 15 | 17 | 19 | 21 | AP#4 | Become up
- by Will Gannett, class of 2002, IB Physics HL
1. How much work is needed to move a -eight.6 m C charge from footing to a point whose potential is +75V?
Well, I know that the equation to utilise is W=qV, where Westward is work in joules, q is accuse in Coul ombs (sp?), and V is voltage in Volts. q is obviously -8.6 m C (-8.6E-six C), and V will be 75, since it has to get from the footing (0V) to the betoken (75V). Since the charge is negative and it is existence moved from 0V to 75V, we know that any work done in this case volition not be work required from an external source, merely work done by the field itself (in the form of electrostatic potential free energy being converted to kinetic free energy). The corporeality of work done will exist W=qV=-8.6E-6*75=-6.45E4J. Over again, this in negative to testify that it is work done by the field rather than piece of work required.
(Tabular array of contents)
3. How much kinetic energy will an electron gain (in joules and eV) if information technology falls through a potential difference of 21,000V in a Boob tube pic tube?
At that place are 2 ways to practise this trouble. 1 volition requite y'all the answer in eV, and ane will give you the respond 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 accuse of one electron, or i.602E-19C. Therefore, the work (which will all be converted to kinetic free energy) is 21000*1.602e-nineteen=3.4E-15J. To get the answer in eV is much easier. Nosotros know that it is one electron being accelerated through 21000V. Therefore, the total number of electron Volts will be 21000, or 21keV.
(Tabular array of contents)
v. How strong is the electric field between 2 parallel plates 5.2mm apart if the potential difference betwixt them is 220V?
Well, you lot should know that there are two different units for electric field force: either North/C, or V/m. These are both in the data bundle, as E=F/q and East=-DV/Dten. E=-DV/Dx is evidently the one to use here.
Eastward=-D5/Dten
E=-220/v.2E-3
E=-42000V/grand
The sign of the answer simply shows the direction of the field.
(Table of contents)
vii. What potential divergence is needed to give a helium nucleus (Q=2e) 65.0keV of KE?
Nosotros know that a unmarried electron, moved through Five volts, volition gain V electron volts of free energy. Similarly, two electrons (or an object with the aforementioned charge as two electrons, such every bit 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 ii electrons (or protons) that has 65000eV of energy would have to only be accelerated through one-half the voltage a single electron would have been. This means that the reply is 65000/ii, or 32500V. To practise the math:
65000eV/2e=32500V.
(Table of contents)
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=.five*m*52. First, we must convert 28.0MeV to joules. This is a uncomplicated matter.
28.0E6eV*1.602e-19C/e
=4.49E-12VC
=4.49E-12J
So, the proton has 4.49E-12 joules of kinetic free energy. Since we know the mass of the proton (1.67e-27kg, for those of you too lazy to actually look at your information packet), information technology is another uncomplicated affair to solve for the velocity.
W=.five*m*52
4.49E-12=.5*1.67E-27*v2
4.49E-12/.five/1.67E-27=vii
v2=5.38E15
And then, v=seven.33E7ms-1
(Tabular array of contents)
15. A +thirty micro coulomb accuse is placed 32 cm from an identical +30 micro coulomb charge. How much piece of work would exist required to motility a +0.50 micro coulomb charge from a point midway between them to t indicate 10 cm closer to either of the two charges?
All nosotros need to exercise is multiply the test accuse by the change in voltage. 5 = Westward/q, so the work, W = Vq where V is the change in voltage, and q is the charge moved.
We now have two locations, ane directly between the ii +30 micro coulomb charges, that is 16 cm from each of them, since they are 32 cm apart, and a second location that is half dozen cm from ane of the charges, and 26 cm from the other. Run into the diagram below:
Then to solve these issues you just do these three steps:
i. Observe the voltage before and subsequently the movement
2. Find the divergence in voltage (subtract)
3. Use W = Vq to calculate the work doneSo the voltage before is simply V = kq/r + kq/r (yay no vectors)
Vbefore = viii.99e9(30e-six)/.sixteen + 8.99e9(30e-6)/.sixteen = 3371250 Five
and after information technology is: (annotation the distances are the merely things that change)
Vafter = 8.99e9(30e-vi)/.06 + 8.99e9(30e-6)/.26 = 5532307.692 Five
The change is that we went up in voltage from 3371250 V to 5532307.692 5, a modify of 2161057.692 volts.
Since we are moving a charge of 0.l micro coulombs through this difference, the work is but
(Table of contents)
W = Vq, W = (2161057.692 V)(0.50e-6 C) = 1.08 J
17. How much voltage must be used to advance a proton (radius 1.2 x 10-15 thousand) so that it has sufficient free energy to penetrate a silicon nucleus? A silicon nucleus has a charge of +14e and its radius is about three.6 10 10-15 m. Assume the potential is that for point charges.
Well, let's kickoff say that the nucleus is stationary and the proton costless. We know that the indicate at which information technology will begin to penetrate is equal to the radius of the silicon nucleus plus the radius of proton. This is i.2E-fifteen+3.6E-15, which is 4.8E-15m. Since the proton is simply able to get this close to the nucleus from kinetic free energy, information technology makes sense that whatever potential energy information technology now has due to proximity must have been obtained from the voltage potential. Since the charge on the proton remains the same and it has no kinetic energy both at the beginning and the stop (it starts out at rest and ends at balance, due to the voltage potential from the accuse of the nucleus), the voltage at those points must exist the aforementioned (i.e. the voltage difference from the get-go to the bespeak of penetration is zippo). What we want to find is the voltage at that point, which will exist the same as the voltage potential through which the proton was accelerated. To do this, we utilize the equation V=kq/r.
V=eight.99e9*14*1.602E-xix/4.8E-15
V=4.2E6=4.2MV
You will discover that the mass and charge of the particle are not relevant. This implies that this would work with blastoff particles equally well, except for the fact that the radius would exist dissimilar. Still, an alpha particle accelerated through that voltage would travel to the same distance from the silicon nucleus.
(Tabular array of contents)
19. Consider indicate a which is 70 cm n of a -3.8 1000 C point accuse, 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 management).
(a) Ok, first nosotros need to find Vb and Va using V=kq/r
Vb=8.99E9*-3.8E-vi/0.8=-42703V
5a=8.99E9*-3.8E-6/0.seven=-48803V
So, Vba=-42703--48803=6100V(b) The electric field is a bit more complicated. Nosotros will use a vector at each point to represent the electrical field strength there and calculate the field strength using E=kq/r2. At point b, E=viii.99E9*-3.8E-vi/0.82=-53378N/C. Since a positive charge placed at betoken b will travel to the right (it is to the left of a negative charge), we know that the vector volition but have an x-component, and it will be positive. So, the field at that bespeak is represented past 53378x+0y. The point b is like. Due east=8.99E9*-3.8E-6/0.72=-69718N/C. The vector at that point volition only have a negative y-component, so the vector volition be 0x+-69718y.
Moving to the next stride, Eastwardb-Ea = (53378-0)x+(0--69718)y=53378x+69718y, in vector component class. (Isn't this neat? Information technology'southward just a chapter 2 problem in disguise!!) The question asks for the answer in angle-magnitude class, though. So, we need to convert. The magnitude is easy. Information technology's just the Pythagorean theorem. 533782+697182=cii, so c=87806N/C. As for the management, tan-1(69718/53378)=53� from vertical, up and to the right. And so in conclusion, the reply is 8.8E4N/C, 53� East of N.
(Table of contents)
21. Ii identical +7.5 m C? signal charges are initially spaced v.5 cm from each other. If they are released at the same instant from rest, how fast will they exist moving when they are very far away from each others? Presume 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 free energy of the system volition equal the corporeality of work required to put the charges in their current state. Nosotros also know that V=kq/r and V=W/q. And then, since kq/r=W/q, W must equal kqii/r.
Due west=kq2/r
West=8.99E9*(7.5E-6)ii/0.055
W=9.194J
So the total energy of the organisation is 9.194 joules. Since both charges take the same mass, and momentum is conserved, they will have equal and contrary velocities. To find these, we use good one-time 1/2*one thousand*v2. Also, because they have the same mass, they will finish upwardly with the same amount of energy, 9.194/2, or four.597J.
one/two*1E-half-dozen*fivetwo=4.597
v2=4.597*2/1E-6
five2=ix.194E6
5=three.0E3m/southward
(Table of contents)
AP # iv goes Here?
<solution goes here>
(Table of contents)
Source: http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/PS17/PS17.htm
0 Response to "what potential difference is needed to give a helium nucleus 65 kev of kinetic energy"
Postar um comentário