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Electrostatics helpby miamiheat5
Tags: electrostatics 
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#1
May1407, 12:40 AM

P: 14

Again I am not ssure what to do.
An electron(mass 9.11x10^31kg) is accelerated in a uniform E field (E=3.0x10^4 N/C) between two parallel charged plates. The separation of the plates is 1.6cm. The electron is accelrated from rest near the negative plate and passes through a tiny hole in the positive plate. With what speed does the electron emerge from the hole? I was thinking you would use qE=ma to find the acceleration, then use kinematics to find the velocity but i am not sure. I would really be happy if anyone can help. 


#2
May1407, 12:56 AM

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Have you tried to solve this based on those assumptions? ;)



#3
May1407, 12:57 AM

P: 14

yes and i got v as being 1.3x10^24m/s



#4
May1407, 12:59 AM

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Electrostatics help
did you remember to take account for the electrons charge?..



#6
May1407, 01:02 AM

P: 14

what do u mean take account for the electrons charge?



#7
May1407, 01:10 AM

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Some books do not give the electron charge in SI units.
Why didnt you post your work done? So we can see were you have done wrong?.. How can one help you if you dont post your work?.. Post your work and I will tell you were it went wrong :) And I also post how I solve this problem. 


#8
May1407, 01:14 AM

P: 14

ok this what i did:
qE=ma 1.602x10^19(3.0x10^4)=(9.11x10^31)a 4.806x10^15=9.11x10^31a a=5.27x10^47 Then: vf^2=vi^2+2ad vf^2=0^2+2(5.27x10^4709.016) sqroot vf^2=sqroot 1.686x10^48 vf=1.3x10^24m/s 


#9
May1407, 01:24 AM

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you shold pratice more on how to calculate with exponents..
4.806x10^15=9.11x10^31a dividing both left and right side with 9.11x10^31 does not give a=5.27x10^47 .. 4.806x10^15 / 9.11x10^31 = 0.528*10^(15+31) 


#10
May1407, 01:29 AM

P: 14

oh ya maybe I should thank you for catching my mistake.
Do you think you can help me with one other problem. With this one I don't have any work because I don't know where to start. A negative charge q is fixed to one corner of a rectangle. What postitive charge must be fixed to corner A and what positive charge must be fixed to corner B, so that the total electric field at the remaining corner is zero? Express your answer in terms of q. Here is the drawing, i will simulate a rectangle without side lines 2d>(should be in middle) A  d>(should be on other side of rectangle) q OB 


#11
May1407, 01:33 AM

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Look at composants in x and y direction. Use coloumbs law and some trigonemtry.
So if the Eletric field should be zero, then the force should be zero too. E = F/q Try from here =) good luck 


#12
May1407, 01:44 AM

P: 14

Ok this I promise will be the last question I need help with again i am not sure how to start the question off.
So this question uses information from: question (7) "An electron(mass 9.11x10^31kg) is accelerated in a uniform E field (E=3.0x10^4 N/C) between two parallel charged plates. The separation of the plates is 1.6cm. The electron is accelrated from rest near the negative plate and passes through a tiny hole in the positive plate. With what speed does the electron emerge from the hole?" The resulting electron beam from question (7) then passes into an evacuated region outside the plates where the E field drops to zero. It then passes through two horizontally oriented defelction plates, the beam travels another 20cm horizontally and strikes a phosphor coated screen. What vertical distance(measured ffrom the path taken if no deflection filed were present) would be measured for the beams defelction?) Is this another question that will use kinematics to be solved? 


#14
May1407, 01:49 AM

P: 14

what formulas will i need to use?



#15
May1407, 02:53 AM

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projectile motion.. i.e free fall combined with transversal motion.
The deflection plates will prevent the particle from avoid its original path, it will not be accelerated by an electric field, and gravity is much much weaker than the effect of the deflection plates. So if you remoce the deflection plates, and dont have any electric field, the electron will just fall beacuse of its mass. 


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