Energy and Work Energy and Work Physics Help Forum Apr 5th 2013, 02:00 PM #1 Junior Member   Join Date: Apr 2013 Posts: 3 Distance ? Good morning/evening <3 For my problem, A metal ball with a mass of 3x10^-6 kg and a charge of +5 mC has a kinetic energy of 6x10^8 j. it's directly at an infinite plane of charge with a charge distribution of +4c/m^2. if it's currently 1m away from the plane of charge, how close will it come to the plane before stopping? (the correct answer is: 0.266 m) ----------------------------------------------------------- (my attemp): 1- finding v2 from Ek = ½mv2 I get v= 20x10^6 m/s 2- F=ma qE=ma a= qE/m 3- find E by : = 2.26x10^11 4- a=3.77x10^14 m/s^s 5- finally: vf2 = vi2 + 2ad d= vi2 /2a = 0.530 m so, 1-0.530 = 0.47m could any one tell me what's wrong?    Apr 5th 2013, 03:08 PM #2 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,344 Given the data you provided, I believe your answer is correct. A way to do this that offers a slight shortcut is this: Work = change in KE, and work = Force x distance, so distance = change in KE/force. Change in KE is 6E+8 Joules Force = Eq = 1.13E+9 N So d= 6E+8 Joules/1.13E+9 N = 0.531 m. And as you say, 1 - 0.531m = 0.469m   Apr 5th 2013, 03:18 PM   #3
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 Originally Posted by Saloom Good morning/evening <3 For my problem, A metal ball with a mass of 3x10^-6 kg and a charge of +5 mC has a kinetic energy of 6x10^8 j. it's directly at an infinite plane of charge with a charge distribution of +4c/m^2. if it's currently 1m away from the plane of charge, how close will it come to the plane before stopping? (the correct answer is: 0.266 m) ----------------------------------------------------------- (my attemp): 1- finding v2 from Ek = ½mv2 I get v= 20x10^6 m/s 2- F=ma qE=ma a= qE/m 3- find E by : = 2.26x10^11.....UNITS? Is it V/m ? 4- a=3.77x10^14 m/s^s 5- finally: vf2 = vi2 + 2ad d= vi2 /2a = 0.530 m so, 1-0.530 = 0.47m could any one tell me what's wrong? The current kinetic energy (when 1 m from the plate) is 6×10^8 J. I don't think you have to find the velocity .. just use Conservation of Energy.

The change of Kinetic Energy is -6×10^8 J when the velocity has fallen to zero.
The change of electric potential is deltaV = E*d (Volts)
so multiply by Q to get the change of potential energy (V*C = J);
Q*E*d = 6×10^8 J
The only unknown is d.

I get d=0.531 m .. same as you .. final position 0.469 m from plate

Have you checked that all the numbers are as given? Why do you believe the correct answer is 0.266 m?
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