- #1
Kumo
- 22
- 1
Hi Everybody,
I am currently battling with the fourth and final question of my A Level physics work for this week. The problem has been driving me crazy, as I seemed to have completed all of the relevant steps, and I find the right answer only with negative exponents, which isn’t much help at all.
It then proceeds to ask questions for the electric force on the α-particles at the surface, and the electric potential of the nucleus, which I was fine with after a few tries. However, question (d) asks the following.
[itex]W=\frac{kqQ}{r}[/itex] or [itex]E= \frac{V}{d}[/itex] together with [itex]V=\frac {kQ}{r}[/itex]
I decided to tackle the problem in two ways, the first is as follows using the electric potential energy formula given above.
[itex]W = \frac{qkQ}{r}[/itex]
[itex]W = \frac{(8.99 * 10^9) * (3.2 *10^{-19}) * (1.44 * 10^{-17})}{(8 * 10^{15})}[/itex]
[itex]W = \frac{(4.143 * 10^{-26})}{(8 * 10^{15})}[/itex]
[itex]W=5.18 * 10^{-12}[/itex] J
This is pretty much the same answer that I found in the back of the book, only with a negative exponent.
My next attempt was a bit of a shot in the dark, but I tried it regardless.
[itex]V = \frac{kQ}{r}[/itex]
[itex]V = \frac{(8.99 * 10^9) * (3.2 * 10^{19})}{(8.0 * 10^{19})}[/itex]
[itex]V = \frac{(2.877 * 10^{-19}}{(8 * 10^{-19}}[/itex]
[itex]V = \frac{3.6 * 10^{-1}}V[/itex]
[itex]E = \frac{V}{d}[/itex]
[itex]E = \frac{(3.6 * 10^{-1})}{(7*10^{-19}}[/itex]
[itex]E = 4.5 * 10^{18} J[/itex] J
I would really appreciate it if someone could shed some light on this issue, as it has really been bugging me. Thank you very much for any help and your time.
I am currently battling with the fourth and final question of my A Level physics work for this week. The problem has been driving me crazy, as I seemed to have completed all of the relevant steps, and I find the right answer only with negative exponents, which isn’t much help at all.
Homework Statement
In a simplified model, a uranium nucleus is a sphere of radius 8.0 * 10-15m. The nucleus contains 92 protons (and rather more neutrons). The charge on a proton is 1.6 * 10-19. It can be assumed that the charge of these protons acts as if it were all concentrated at the centre of the nucleus. The nucleus releases an α-particle containing two protons (and two neutrons) at the surface of the nucleus.
It then proceeds to ask questions for the electric force on the α-particles at the surface, and the electric potential of the nucleus, which I was fine with after a few tries. However, question (d) asks the following.
Calculate: the electric potential energy of the α-particle when it is at the surface of the nucleus.
Homework Equations
[itex]W=\frac{kqQ}{r}[/itex] or [itex]E= \frac{V}{d}[/itex] together with [itex]V=\frac {kQ}{r}[/itex]
The Attempt at a Solution
I decided to tackle the problem in two ways, the first is as follows using the electric potential energy formula given above.
[itex]W = \frac{qkQ}{r}[/itex]
[itex]W = \frac{(8.99 * 10^9) * (3.2 *10^{-19}) * (1.44 * 10^{-17})}{(8 * 10^{15})}[/itex]
[itex]W = \frac{(4.143 * 10^{-26})}{(8 * 10^{15})}[/itex]
[itex]W=5.18 * 10^{-12}[/itex] J
This is pretty much the same answer that I found in the back of the book, only with a negative exponent.
My next attempt was a bit of a shot in the dark, but I tried it regardless.
[itex]V = \frac{kQ}{r}[/itex]
[itex]V = \frac{(8.99 * 10^9) * (3.2 * 10^{19})}{(8.0 * 10^{19})}[/itex]
[itex]V = \frac{(2.877 * 10^{-19}}{(8 * 10^{-19}}[/itex]
[itex]V = \frac{3.6 * 10^{-1}}V[/itex]
[itex]E = \frac{V}{d}[/itex]
[itex]E = \frac{(3.6 * 10^{-1})}{(7*10^{-19}}[/itex]
[itex]E = 4.5 * 10^{18} J[/itex] J
I would really appreciate it if someone could shed some light on this issue, as it has really been bugging me. Thank you very much for any help and your time.
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