The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates?

Question

The vertical deflecting plates of a typical classroom oscilloscope consist of a pair of parallel square metal plates that carry equal but opposite charges. The typical dimensions of these plates are approximately 3.0 cm on each side, with a separation of about 4.7 mm. The plates are positioned close enough that we can disregard any fringing effects at the edges.

Under these conditions, how much charge is on each plate?  
What is the strength of the electric field between the plates?  
If an electron is ejected at rest from the negative plate, how fast is it moving when it reaches the positive plate?

Please assist me with these questions using only the data provided.

Anonymous · Accepted Answer

The voltage across the vertical deflection plates of a typical classroom 
 cathode ray oscilloscope is around 150 volts.
 1- In order to determine the charge on each plate we must calculate
     the capacitance, C, of the plates and then find Q = CV.
 C = εoA/d = (8.85×10^-12)0.03²/4.7×10-³
 C = 16.95×10^-13 F = 1.69 pF
 Q = CV = 150*1.69×10^-12 = 0.25 nC
 2-  E = V/d = 150/4.7×10-³ = 3.2×10^4 N/C
 3- qV = ½ mv²,  
     v = √{2qV/m} = √{2(1.6×10^-19)150/9.1×10^-31} = 7.263×10^6 m/s

Anonymous · Answer

It seems like the only way to help you is to tell you why the data is insufficient.
All you have given is the area and separation of the plates. From this we can calculate the capacitance (C = e0*area/separation). But the answers you seek simply can’t be found from this information. What’s needed is something to indicate the voltage, the charge or the field. 
It’s analogous to asking, how much liquid is in a bottle whose base is 0.1 m^2? How does the pressure change with depth? Ignoring drag, if you release a bubble at the bottom, how fast is it going when it reaches the top? Get it?

Anonymous · Answer

you probably did no longer point out the voltage V utilized to the plates. if V is that voltage, then a) cost on the plates Q = capacitance C x V = eo A V / d = 9×10^-12 x (3.2×10^-2)^2 V / 5.3×10^-3 ( coulomb) b) electric field E = V / d = V / 5.3 x10^-3 volt / m c) if ‘v’ is the cost of elctron, then a million/2 m v^2 = V e, so as that v = squre root of ( 2 V x e / m ) = squre root of ( 2 V x a million.76×10^11 ) m/sec so, all those values might nicely be calculated utilising V cost. o.ok.stable luck.

Josefa Kozey MD

The vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates?

3 Answers

AI-Powered Insights

Want to answer this question?

Categories

Tags

Related Questions