Determination of charge/mass value in E field.

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To determine the charge-to-mass ratio (q/m) of a particle deflected in an electric field, the electric field strength is calculated as 600,000 Vm^-1 based on a potential difference of 3000V over a distance of 5mm. The time the particle spends between the plates is approximately 1.03 X 10^-7 seconds. The vertical acceleration of the particle can be expressed as a function of q/m, leading to the equation a = 600,000(q/m). The horizontal velocity remains constant at 3.9 X 10^6 m/s, while the vertical velocity is derived from the acceleration multiplied by time. The challenge lies in isolating q/m from the equations, as they contain it on both sides, complicating the calculation.
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I've been trying this out many times I just couldn't get the answer, I hope someone could help me out here.

A stream of nuclear particles are traveling horizontally at 3.9 X 10^6 ms^-1 are deflected as it passes through a vacuum between 2 parallel plates which are 5mm apart and 40mm in length. The P.D between the plates is 3000V and the angle of deflection is 4.3 degrees. Calculate q/m of a particle.

My working:

I first found out the field strength between the plates: dV/dr = 3000/0.005 = 600000 Vm^-1.

Then the time for which a particle is between the plates: 0.4/3.9 X10^6 = 1.0256 X 10^-7 s.

To find acceleration of a particle vertically downwards in terms of q/m would be => qE = ma, => a=qE/m = 600000q/m.

Horizontal velocity is after it leaves the E field would still be 3.9 X 10^6, while vertical velocity is vertical acceleration multiplied by the time, => (6.15 X 10^-3)(q/m) ms^-1.

I've found all I could, how can I solve for q/m? I tried using the equations for linear acceleration but they all consists of q/m on both sides which I cannot make the subject to determine the value.

Any help much appreciated.

alvin
 
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