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upward-pointing electric field `vecE` is located between two parallel plates that have...
upward-pointing electric field `vecE` is located between two parallel plates that have potential difference `5 xx 10^3` V. The plates have length L=10cm and separation d=2 cm. A particle with a mass m=`2,5xx10^-10` kg and a positive charge q= `+3xx10^-13` C enters the region between the plates, initially moving along the plates with speed of 20m/s. Assume that the effect of gravity can be neglected
1. what is the magnitude of the electric field `vecE` between the plates ?
2. find the magnitude of electrostatic force on the particle!
3. what is the vertical deflection Y of the particle at the far edge of the plates?
1 Answer | add yours
1) The magnitude of electric field between the plates is
`E = (Delta V)/d` , where `Delta V` is the potential difference between the plates
`Delta V = 5 * 10 ^3 V` and d is the separation: d = 2 cm = 0.02 m
So `E = (5*10^3)/0.02 =2.5*10^5 V/m`
2) The magnitude of force on the particle is `F = |q|*E` , where q is the electric charge.
`F = 3*10^-13 * 2.5 *10^5 V/m=7.5*10^-8 N`
3) The vertical deflection is determined by the acceleration a in the vertical direction due to the electric force, according to the 2nd Newton's Law:
ma = F where m is the mass of the particle.
`a=F/m = (7.5*10^-8)/(2.5*10^-10) = 300 m/s^2`
Since the initial velocity has no vertical component, the vertical deflection is
`y = (a*t^2)/2` , where t is the time it takes for the particle to reach the end of the plates.
Time can be found from the horizontal component of the motion, since horizontal component of the particle's velocity is constant throughout:
`L = V_0*t`
`t = L/V_0 = (0.1 m)/(20 m/s)=0.005 s`
Then the vertical deflection is `y = (300*(0.005)^2)/2 = 3.75*10^-3 m = 3.75 mm`
Posted by ishpiro on July 2, 2013 at 3:15 PM (Answer #1)
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