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

Question

Deflecting Plates of an OscilloscopeThe vertical deflecting plates of a typical classroom oscilloscope are a pair of parallel square metal plates carrying equal but opposite charges. Typical dimensions are about 3.0 cm on a side, with a separation of about 4.7 mm. The plates are close enough that we can ignore fringing at the ends.1- Under these conditions. how much charge is on each plate?2- How strong is the electric field between the plates? 3- f an electron is ejected at rest from the negative plates, how fast is it moving when it reaches the positive plate?plz help me this is all the data given {plz don’t tell me it’s not enough},

Abbie Toy · 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

Abbie Toy · 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?

Abbie Toy · 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.

Mr. Ellis Hintz

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

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