Posted 13 November 2005 - 08:22 PM
Can someone pelase help with this problem?
Electrons are accelerated through a large potential difference of 7.5x10^5V. The electronas are initially at rest. Calculate the speed reached by these electrons.
Wouldn't you need more information?
I thought about writing just under the speed of light but there is another quesiron like this.
Thanks
Ian
Posted 13 November 2005 - 08:50 PM
OK, I am remembering the constants from memory from several months ago, so correct them if they are wrong. Also, the final answer to this is very high so either I have done something silly or you will need to consider relativistic effects.
Posted 13 November 2005 - 09:11 PM
Why i there a maximum energy the electron can reach? Why would it not accelerate from the speed you calculated
Posted 13 November 2005 - 10:04 PM
Well according to your equation QV = 1/2mv

, the accelerating force IS the electric field due to the potential difference, me thinks.
Posted 13 November 2005 - 10:06 PM
Well from what I did the electron would end up faster than c and obviously that isn't going to happen, as as v tends to c, m tends to infinity, etc.
That's why I'm suggesting that either you need to consider relativistic effects or I have done something stupid.
Posted 13 November 2005 - 10:15 PM
But why does the potential difference matter? if it was 1V of 1,000,000V wouldn't it never reach a maximum speed. Wouldn't it jsut get getting closers to C?
Posted 13 November 2005 - 10:31 PM
yeah if i remember right didnt we just say that although that was the value found from calculation this would never be the case becase v cannot be greater than c
although i think most time i got that it was due to a stupid mistake
Posted 19 January 2006 - 03:25 PM
![\[
\begin{array}{l}
QV = \frac{1}{2}mv^2 \\
v = \sqrt {\frac{{2QV}}{m}} \\
v = \sqrt {\frac{{2 \times \left( {1.6 \times 10^{ - 19} } \right) \times \left( {7.5 \times 10^5 } \right)}}{{9.11 \times 10^{ - 31} }}} \\
\end{array}
\]
\[
\begin{array}{l}
QV = \frac{1}{2}mv^2 \\
v = \sqrt {\frac{{2QV}}{m}} \\
v = \sqrt {\frac{{2 \times \left( {1.6 \times 10^{ - 19} } \right) \times \left( {7.5 \times 10^5 } \right)}}{{9.11 \times 10^{ - 31} }}} \\
\end{array}
\]](/latexrender/pictures/3f3a17fd91a1794bf94c6577e106c7f4.gif)
[right]

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Where did you get these math graphics from that i cant find anywhere how to properly make.
Posted 19 January 2006 - 05:03 PM
QUOTE(Jason Bourne @ Jan 19 2006, 03:25 PM)
Where did you get these math graphics from that i cant find anywhere how to properly make.


I posted a quick introduction here:
http://www.hsn.uk.net/forum/index.php?show...indpost&p=65413