Jump to content

  • You cannot start a new topic
  • You cannot reply to this topic

Quick Quesiton

Ian

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

werlop

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.


\[
\begin{array}{l}
 QV = \frac{1}{2}mv^2  \\ 
 v = \sqrt {\frac{{2QV}}{m}}  \\ 
 v = \sqrt {\frac{{2 \times \left( {1.6 \times 10^{ - 19} } \right) \times \left( {7.5 \times 10^5 } \right)}}{{9.11 \times 10^{ - 31} }}}  \\ 
 \end{array}
\]

Ian

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

dfx

Posted 13 November 2005 - 10:04 PM

Well according to your equation QV = 1/2mv power2.gif , the accelerating force IS the electric field due to the potential difference, me thinks.

werlop

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.

Ian

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?

Dave

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

Jason Bourne

Posted 19 January 2006 - 03:25 PM



\[
\begin{array}{l}
 QV = \frac{1}{2}mv^2  \\ 
 v = \sqrt {\frac{{2QV}}{m}}  \\ 
 v = \sqrt {\frac{{2 \times \left( {1.6 \times 10^{ - 19} } \right) \times \left( {7.5 \times 10^5 } \right)}}{{9.11 \times 10^{ - 31} }}}  \\ 
 \end{array}
\]
[right]View Post[/right]
[/quote]

Where did you get these math graphics from that i cant find anywhere how to properly make. rolleyes.gif

George

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. rolleyes.gif

View Post


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

  • You cannot start a new topic
  • You cannot reply to this topic