A particle possessing 3.25x10^-6 C of charge and mass 0.00455 kg is fired at a s
ID: 2062251 • Letter: A
Question
A particle possessing 3.25x10^-6 C of charge and mass 0.00455 kg is fired at a speed of 3.57 m/s through two charged plates of length 0.245 m, as shown in the figure. If the electric field is constant at 2750 N/C between the two plates* and directed upwards, calculate the distance y in which the charge falls below its intended path.
_____________ meters
Which field strength will allow the particle to pass between the plates along a straight path.
_______________ N/C
*In actuality, the electric field will vary considerable near the edges of the plates. We will ignore this situation and consider the electric field constant everywhere between the plates.
Explanation / Answer
Time to cross the field = 24.5/357 = 0.0686 s F on charge = 3.25*10^(-6) *2750 a = 1.964 m/(s*s) so y(upwards) = 0.5*1.964*0.0686*0.0686 = 4.621 * 10^(-3) m (Ans) An opposite field of 2750 N/C will allow the charge to pass in straight line
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