What is the instantaneous potential?

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In summary, the potential difference between plates A and B is not large enough to cause the electron to reach plate B.
  • #1
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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
1608128633908.png

##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';
 
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  • #2
Not necessarily. When the reverse voltage is applied, the electron already has a velocity in the upward direction. How far will it travel before the direction of the velocity is reversed?
 
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  • #3
Ok, Thank you for the reply. Again when the reverse voltage of 50V is applied
##\frac{mv_0^2} 2 - qV##
##v_0 = 5906 m/s## from the previous calculations
##\frac{mv_0^2} 2 = KE = 1.58*10^{-23}J ##
##qV = PE = 8*10^{-18} J ##
since the PE is greater than KE, it will not reach plate "B". Am I correct?
 
  • #4
PhysicsTest said:
##qV = PE = 8*10^{-18} J ##
since the PE is greater than KE, it will not reach plate "B". Am I correct?
It seems that you used 50 V for ##V##. That is the potential difference between plates A and B. Is that the potential difference you want in the equation?

On edit: Your numbers in part (a) are tangled up because you chose to mix numbers and symbols. I very strongly recommend that you start over again using symbols only. Let ##\alpha = 10^7 ~\mathrm{V/ s}## and ##d=0.03 ~\mathrm{m}## and write the acceleration as ##a=\dfrac{q\alpha t}{md}.## Proceed as before and substitute the numbers at the very end. The same applies for the other parts. They will be easier to troubleshoot if you have symbolic expressions especially if you use results from part (a) which is the case here.
 
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  • #5
Let me do it from beginning, the variables are
##m_e -> 9.1*10^{-31}kg, ## ##q_e -> 1.6*10^{-19}C, ##
##\alpha -> 10^7 V/s, ##
##d_t -> 0.03m (distance between plates),##
## a->acceleration, m/s^2##
##v-> velocity, m/s ##
##t -> time, s ##
##a=\frac {q_e\alpha t} {d_t m_e}##
##\frac {dv} {dt} = \frac {q_e\alpha t} {d_t m_e} ##
##dv = \frac {q_e\alpha t dt} {d_t m_e } ##
##v = \frac {q_e\alpha t^2} {2d_t m_e } + C1 ## ; when t=0; v=0 Hence C1=0;
##v = \frac {q_e\alpha t^2} {2d_t m_e } m/s ---> eq1##
##dx = \frac {q_e\alpha t^2dt} {2d_t m_e }##
##x=\frac{q_e\alpha t^3} {6d_t m_e} + C1## C1 is 0 as x=0 at t=0;
##x=\frac{q_e\alpha t^3} {6d_t m_e} ## ---> eq2
Expressing "t" in terms of "x"
## t^3= \frac{6d_tm_e x} {q_e\alpha} ##
##t = (\frac{6d_tm_e x} {q_e\alpha})^{\frac 1 3} ##
When the electron has traveled a distance of 0.028m, what is "t" substituting
##x = 0.028m## what is "t"
##t=1.47 * 10^{-7} s##. ---> eq3
The potential when electron reaches 0.028m is
##V = \alpha * t ##
##V = 10^7*1.47*10^{-7} V##
##V=1.47V## ----> eq4 (Answer part "a" when electron reaches point "D")
1608354707503.png


The time to reach point "D" is as per eq3 is
##t=1.47 * 10^{-7} s##
The velocity at point "D" is as per eq1 is
##v_d = 6.3*10^5 m/s##

b. To find if the electron reaches point "B", check if the velocity of electron at the time it reaches plate "B" is positive.
Energy at point D = Energy at Plate B;
##KE_d + PE_d = KE_b + PE_b ##
##PE_d = q_e*1.47V ## derived for part a
##PE_b ## = 0V
##KE_d = 0.5mv_d^2##
##KE_b = ?##
##KE_b = 0.5mv_d^2 - q_eV_d##
##V_d=1.47V## derived from eq4. Substituting the values
##KE_b = 1.8*10^{-19} J##
##PE_b = 2.35*10^{-19} J ##
Since the ##PE_b## is greater than ##KE_b## the electron does not hit the plate B.
 
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  • #6
When i rethink part b, i feel i made some mistakes, when i apply the reverse potential the voltage at point "B" should be 50V not 0, since the KE is decreasing. If i re modify and apply the energy conservation again.
##KE_d + PE_d = KE_b + PE_b ##
##0.5mv_d^2 + qV_d = 0.5mv_b^2 = qV_b##
##V_d = 1.47V, v_d = 6.3*10^5, V_b = 50V , v_b=? ##
##KE_d = 1.8*10^{-19}, PE_b = 2.35*10^{-19} , PE_b = 8*10^{-18}##
##1.8*10^{-19} + 2.35 * 10{-19} = 8*10^{-18} + KE_b##
##KE_b + 80*10^{-19} = 4.15*10^{-19}##
I am confused which is correct? Sometimes i think of applying ##qE = ma ## find the time, velocity to cover 0.03- 0.028 = 0.002m. Which method should i follow?
 
  • #7
Note ##1.42\times10^{-7}s=0.142μs## not ##14.2μs## (post #1). I haven’t checked anything else.

You have few replies partly because your working is very difficult to read (gives me a headache!). Here are some hints:

If you have to include text inside a Latex expression use \text{} to avoid losing the spaces. E.g.
##this is a short sentence ## (done without \text{})
##\text{this is a short sentence}## (done with \text{})

Include some blank lines where appropriate, to improve readability (like we use paragraphs and blank lines for text).

Avoid unnecessary clutter, e.g. don’t using ##d_t## for a fixed distance when ##d## would do.

In part b), you want to see if the electron reaches plate B. Here’s a strategy:

Imagine standing on a chair in a room. You throw a ball upwards. You know the initial speed, the launch height, the ceiling height and the acceleration (g). How would you calculate:
i) whether or not the ball hits the ceiling?
ii) if the ball doesn’t hit the ceiling, how long it takes to reach the ground?
iii) if the ball doesn’t hit the ceiling, what is its speed when it hits the ground?
 
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  • #8
Yes, i understand i try if i can improve.
##v_b^2 = v_d^2 - 2ad; d= 0.002m, v_d= 6.3*10^5 m/s##
##a = \frac{qV} {md} ; V=50 V, d=0.002 m##
## a = 4.39*10^{15} m/s^2##
##0 = 3.9*10^{11}-2*4.39*10^{15}*x##
##x = 4.4*10^{-5} m < 0.002 m ## it does not touch plate "B".
 
  • #9
You are on the right track. But the electric field is not 50/0.002 Between what positions is the 50V is being applied??
 
  • #10
Steve4Physics said:
You are on the right track. But the electric field is not 50/0.002 Between what positions is the 50V is being applied??
Not sure if i am correct. The potential at point D from post #5 is ##V_d = 1.47 V## and ##V_b = 50 V##, ##V_{bd} = 48.53V##. Hence ##E = \frac {48.53} {0.002} = 24265 V/m##. Am i correct?
 
  • #11
No. The electric field between the plates is constant. It has magnitude given by ##E=\frac{V}{d}## where ##d## is the separation between plates. After the reversal, the direction of the electric field is from high to low potential, here from A to B. The expression for the potential is ##V=Ex+V_0## where ## V_0## is a constant chosen to be the potential at ##x=0##. Here you can pick ##V_0=0##. The potential is not as relevant as the acceleration for answering parts (c) and (d). Since you need the time of flight and the impact velocity, you can get both using the appropriate kinematic equations.

I repeat my exhortation to use symbols instead of numbers. It will make our job of helping you much, much, much easier and is in the interests of all involved.
 
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  • #12
PhysicsTest said:
Not sure if i am correct. The potential at point D from post #5 is ##V_d = 1.47 V## and ##V_b = 50 V##, ##V_{bd} = 48.53V##. Hence ##E = \frac {48.53} {0.002} = 24265 V/m##. Am i correct?
Not correct. In your original question you are told:
"When the electron is 2.8cm from the bottom plate, a reverse voltage of 50V replaces the linearly rising voltage. "

EDIT: Note the use of the word 'replace'. The original rising voltage has gone!

If you wanted to measure the 50V, where would you connect the voltmeter?
 
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  • #13
The value for ##v_d## is a bit too high. It looks like in Equation 1, post #5 you used dt = 0.028 m instead of 0.03 m.
 
  • #14
Steve4Physics said:
Not correct. In your original question you are told:
"When the electron is 2.8cm from the bottom plate, a reverse voltage of 50V replaces the linearly rising voltage. "
EDIT: Note the use of the word 'replace'. The original rising voltage has gone!
If you wanted to measure the 50V, where would you connect the voltmeter?
Ok, now i understand the question clearly. The E field is ##E = \frac{V} {d} = \frac{50} {0.03} = 1.66KV##
kuruman said:
The value for ##v_d## is a bit too high. It looks like in Equation 1, post #5 you used dt = 0.028 m instead of 0.03 m.
I have re verified, i have taken the value of ##d_t = 0.03 m##.
##v_b^2 = v_d^2 - 2ax; v_d= 6.3*10^5 m/s##
##a = \frac{qV} {md} ; V=50 V, d=0.03 m##
## a = 2.93*10^{14} m/s^2##
##0 = 3.9*10^{11}-2*2.93*10^{14}*x##
##x = 6.65*10^{-4} m < 0.002 m ## it does not touch plate "B".
 
  • #15
PhysicsTest said:
Ok, now i understand the question clearly. The E field is ##E = \frac{V} {d} = \frac{50} {0.03} = 1.66KV##

I have re verified, i have taken the value of ##d_t = 0.03 m##.
##v_b^2 = v_d^2 - 2ax; v_d= 6.3*10^5 m/s##
OK, you may have re-verified, but now that you have posted in symbolic form the equation you used, I can see your error. You did not use Equation 1 in post #5 as per your claim. You used the kinematic equation that is valid only when the acceleration is constant. In the first leg of the trip the acceleration is not constant but depends on time and you have already derived the expression for that. Note that mechanical energy is not conserved during the first leg of the trip because the potential energy is time-dependent. That's another way of saying that you cannot use the equation you used.

Use Equation 1 in post #5 to find the velocity.


Sorry, I take it all back. I got confused about the meaning of ##v_d##. The number is correct.
 
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  • #16
PhysicsTest said:
Ok, now i understand the question clearly. The E field is ##E = \frac{V} {d} = \frac{50} {0.03} = 1.66KV##
What units are used for electric field? What is the difference between 'k' and 'K' in the context of units?
PhysicsTest said:
I have re verified, i have taken the value of ##d_t = 0.03 m##.
##v_b^2 = v_d^2 - 2ax; v_d= 6.3*10^5 m/s##
##a = \frac{qV} {md} ; V=50 V, d=0.03 m##
## a = 2.93*10^{14} m/s^2##
##0 = 3.9*10^{11}-2*2.93*10^{14}*x##
##x = 6.65*10^{-4} m < 0.002 m ## it does not touch plate "B".
I guess you are taking the +x direction to be plate A to plate B. The 'reverse voltage' (50V) causes an acceleration in the -x direction. So you should have a negative value for acceleration if you are applying the kinematics equations consistently.
 
  • #17
Now it is more confusion. The sequence of steps i followed
Step1: When accelerating potential is applied till 0.028m, i used the Equation 1 in post #5 to calculate the velocity ##v_d##.
Step2: When the electron reaches the 0.028m, the reverse voltage is applied, kinematic equation is used in post #14, since the electron has another ##0.03 - 0.028 = 0.002 m## to travel if it has to hit the plate "B" and constant reverse acceleration is applied. So, i used the initial velocity of ##v_d## and it is decelerating and i am finding the distance it can travel. What is the mistake?
What units are used for electric field? What is the difference between 'k' and 'K' in the context of units?
I am not sure how i am missing writing the units. Sorry the unit of Electric field is kV/m basically i should have written ##E = -1.66*10^3 V/m ##
The original kinematic equation is ## v_f^2 = v_i^2 + 2ad; v_f=v_b = 0, v_i =v_d ##
##v_b^2 = v_d^2 - 2ax; v_d= 6.3*10^5 m/s##
##a = -\frac{qV} {md} ; V=-50 V, d=0.03 m##
## a = -2.93*10^{14} m/s^2##
##0 = 3.9*10^{11}-2*2.93*10^{14}*x##
##x = 6.65*10^{-4} m##
 
  • #18
Steve4Physics said:
What units are used for electric field? What is the difference between 'k' and 'K' in the context of units?

I guess you are taking the +x direction to be plate A to plate B. The 'reverse voltage' (50V) causes an acceleration in the -x direction. So you should have a negative value for acceleration if you are applying the kinematics equations consistently.
I think the equation ##0 = 3.9*10^{11}-2*2.93*10^{14}*x## is correct. In symbolic form it is a rearrangement of ##2(-a)\Delta x_{\text{max}}=0^2-v_d^2## in which the one-dimensional dot product betwen the acceleration and the displacement is a negative number and so is the change in speed squared on the other side. It's the work-energy theorem in disguise.
 
  • #19
PhysicsTest said:
Now it is more confusion. The sequence of steps i followed
Step1: When accelerating potential is applied till 0.028m, i used the Equation 1 in post #5 to calculate the velocity ##v_d##.
Step2: When the electron reaches the 0.028m, the reverse voltage is applied, kinematic equation is used in post #14, since the electron has another ##0.03 - 0.028 = 0.002 m## to travel if it has to hit the plate "B" and constant reverse acceleration is applied. So, i used the initial velocity of ##v_d## and it is decelerating and i am finding the distance it can travel. What is the mistake?

I am not sure how i am missing writing the units. Sorry the unit of Electric field is kV/m basically i should have written ##E = -1.66*10^3 V/m ##
The original kinematic equation is ## v_f^2 = v_i^2 + 2ad; v_f=v_b = 0, v_i =v_d ##
##v_b^2 = v_d^2 - 2ax; v_d= 6.3*10^5 m/s##
##a = -\frac{qV} {md} ; V=-50 V, d=0.03 m##
## a = -2.93*10^{14} m/s^2##
##0 = 3.9*10^{11}-2*2.93*10^{14}*x##
##x = 6.65*10^{-4} m##
There is no mistake. I got confused with all the numbers and what they mean. Please see post #15. I think you are good to move on to parts (c) and (d).
 
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  • #20
kuruman said:
There is no mistake. I got confused with all the numbers and what they mean. Please see post #15. I think you are good to move on to parts (c) and (d).
I agree with kuruman - I think you are correct so far (but your working is difficult to check/follow).

A few points to be aware of:

'k' means 'kilo' (1000) and 'K' means 'kelvin' (a unit of temperature ). It's important to not mix them up (but many people incorrectly use 'K' for 'kilo').

V/m is equivalent to N/C. Either can be used for electric field.

You can often solve a problem using different methods. For example, if you want to know how high you can throw a ball given the initial velocity and g, you can use a kinematics equation or energy conservation. They must give the same answer. You have to decide which method is most convenient.
 
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Related to What is the instantaneous potential?

What is the instantaneous potential?

The instantaneous potential is a term used in physics to describe the potential energy of a system at a specific moment in time. It is a measure of the energy stored in a system due to the position of its components and can be calculated using mathematical equations.

How is the instantaneous potential different from the total potential?

The instantaneous potential refers to the energy of a system at a specific moment, while the total potential refers to the total energy of a system at any given time. The total potential takes into account the potential energy at all points in time, while the instantaneous potential only considers the energy at one point in time.

What factors affect the instantaneous potential?

The instantaneous potential is affected by the position and configuration of the components of a system. It also depends on the forces acting on the system and the properties of the objects within the system, such as mass and charge.

How is the instantaneous potential related to the concept of potential energy?

The instantaneous potential is a measure of potential energy at a specific moment in time. It is related to the overall concept of potential energy, which is the energy an object has due to its position or configuration in a system. The instantaneous potential is a more precise and specific measurement of potential energy at a particular point in time.

Can the instantaneous potential change over time?

Yes, the instantaneous potential can change over time as the position and configuration of a system's components change. This change in potential energy can be caused by external forces acting on the system or by the system's internal dynamics. In some cases, the instantaneous potential may remain constant if the system is in a state of equilibrium.

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