Recent content by 1126
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Electric force acting on an electron
OH! That makes so much more sense! Thank you so much for explaining that! That was so helpful! I really appreciate it! :)- 1126
- Post #7
- Forum: Introductory Physics Homework Help
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Electric force acting on an electron
I do not know the size of the Electric Field strength. :/ that's why I resorted to the fundamental charge of 1.60 x 10^-19.- 1126
- Post #5
- Forum: Introductory Physics Homework Help
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Find the Electric Field at a midway point
Thank you so so so much! that was so helpful! I think I solved it correctly now. :) thanks so much!- 1126
- Post #5
- Forum: Introductory Physics Homework Help
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Electric force acting on an electron
Well, I'm finding the Electric Force which could also be E=F/q. Although, without any mass, I can't solve for F and put it into the equation. You were correct with the 8.64 x 10^-16, I simply calculated it incorrectly. So, is the way I went about solving this equation correct?- 1126
- Post #3
- Forum: Introductory Physics Homework Help
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Electric force acting on an electron
An elctron moving through an electric field experiences an acceleration of 5.4 x 10^3 m/s^2. Find the electric force acting on the electron. I think the equation F=ma may be used and maybe E=F/qsubo Do I need to use the fundamental charge to find an answer? using the fundamental charge...- 1126
- Thread
- Electric Electric force Electron Force
- Replies: 6
- Forum: Introductory Physics Homework Help
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Find the Electric Field at a midway point
I still don't really understand. Is .3m really the right radius? It's the distance between the charges, but I'm not sure if it's the right radius to use. Any suggestions? :/- 1126
- Post #3
- Forum: Introductory Physics Homework Help
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Find the Electric Field at a midway point
The problem is: Find the electric field at a point midway between two charges of +30.0 x 10^-9C and +60.0 x 10^-9 C separated by a distance of 30 cm. I used E=k (Q/r^2) My attempt is: E=(9x10^9)(30.0 x 10^-9C)/(.3m^2) E=3000 N/C E=(9x10^9)(60.0 x 10^-9 C)/(.3m^2) E=6000 N/C...- 1126
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- Electric Electric field Field Point
- Replies: 4
- Forum: Introductory Physics Homework Help