Cathode ray tube(CRT) question

[F|234K]

in a CRT, the deflection on the screen is 2.4 cm when the accelerating voltage is 480 V and the deflecting voltage is 36 V. what deflection on the screen will you see if the accelerating voltage is 960 V (doubled) and the deflecting voltage is 18 V (halfed)?

answer is 0.6 cm

can anyone please explain how to do this qeustion. (i tried, but there are so many formulas to plug and it's very complicated, one have to keep track of every change in the speed, then time, then distance, then finally the deflection. i am sure there's a better way to do it, like using a ratio, but i have no idea how to do it...)

 

[F|234K]

no one has any idea?

 

[whozum]

You don't really need to plug into formulas, its just a matter of ratios.
It's going twice as fast, so the deflecting voltage has half the effect that it normally would if it were kept constant (since the electron gets there faster). On top of that, the deflecting votlage is halved, which makes the deflecting force half as weak and the electron is deflected less.
Half the time to get deflected + Half the deflecting force = Quarter deflecting distance.

 

[F|234K]

hold on...why the electron moves twice as fast, shouldn't it be squarroot(2) times as fast?

equation used: 1/2mv^2=Vq=KE

if V is doubled, and m and q is constant, then v should be squarroot(2) times as fast, isn't it?

 

[F|234K]

can anyone please give some idea......i am sure there are a lot of people who can help me...it's not that hard is it?

 

[seiferseph]

the deflection in the crt (provided everything else stays constant) is equal to the deflecting voltage over the accelerating voltage. that is D = Vd/Va. simplifying equations and solving with ratios will get you the 0.6 m.

 

[F|234K]

thanks a lot seiferseph. how u figure the equation D=Vd/Va?...whats the physics behind it? or it's simply using the rule of ratio?

 

[jdavel]

seiferseph's answer was essentially right

whozum's explanation was close, but you're right about the forward velocity decreasing by sqrt(2), not 2.

The electron passes all the way through the accelerating voltage, so halving this voltage reduces forward speed by sqrt(2). This means the electron spends sqrt(2) times as long in the deflecting field. Since this field is twice as strong, the acceleration, while the electron is in the field is doubled. Since v =at, you get an total increase in deflection speed of 2*sqrt(2). The reduction in forward speed, already accounted for, gives you the other sqrt(2) factor.