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PhysicsTest

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- Homework Statement
- An electron is released with zero initial velocity from the lower of a pair of horizontal plates which are 3cm apart. The accelerating potential between these plates increases from zero linearly with time at the rate of 10V/uS. When the electron is 2.8cm from the bottom plate, a reverse voltage of 50V replaces the linearly rising voltage.

a, What is the instantaneous potential between the plates at the time of the potential reversal?

b, With which electrode does the electron collide?

c, What is the time of flight?

d, What is the impact velocity of the electron?

- Relevant Equations
- F=qE;

From the previous posts help i am able to understand make some progress, thank you very much. I am trying only the part a of the problem, once it is ok i try c and d, the problem is I do not have answers, so i cannot confirm if my working is correct or wrong.

a. ##ma = qE## -> eq1

##V1 = 10t ## t in ##\mu s##

Using eq1

##ma = \frac {10qt} {0.03} => m \frac{dv} {dt} = 333.3qt##

##dv = \frac{333.3q} {m} t dt## Integrating

##v = 166.6 \frac q m t^2 + C ## when t=0 v=0; Hence C=0;

##dx = 166.6 \frac q m t^2 dt##

##x = 55.5 \frac q m t^3 + C1 ## when t=0; C1=0;

when x = 0.028m; t=?

##t =14.2\mu s ## The voltage is

##Voltage =10*14.2V= 142V##

b. For this i did not think much, since reverse voltage is applied, the electron should hit plate 'A';

a. ##ma = qE## -> eq1

##V1 = 10t ## t in ##\mu s##

Using eq1

##ma = \frac {10qt} {0.03} => m \frac{dv} {dt} = 333.3qt##

##dv = \frac{333.3q} {m} t dt## Integrating

##v = 166.6 \frac q m t^2 + C ## when t=0 v=0; Hence C=0;

##dx = 166.6 \frac q m t^2 dt##

##x = 55.5 \frac q m t^3 + C1 ## when t=0; C1=0;

when x = 0.028m; t=?

##t =14.2\mu s ## The voltage is

##Voltage =10*14.2V= 142V##

b. For this i did not think much, since reverse voltage is applied, the electron should hit plate 'A';