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zimovet [89]
3 years ago
13

A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg

ative plate with a speed of 55000m/s . What will be the final speed of an electron released from rest at the negative plate?
Physics
2 answers:
Natali5045456 [20]3 years ago
5 0

Answer: Vf = 2,400,000 m/s

Explanation:

1) The only relevant force is the electrostatic force

2) The formula for the electrostatic force is F = E×q

Where E is the electric field and q is the magnitude of the charge.

3) Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magntitude of the electric forces acting in both proton and electron are the same

4) Fe = Fp (Fe stands for force on the electron and Fp stands for force on the proton).

5) Using second law of Newton, Force = mass × acceleration

Fe = Me × Ae (Me is mass of the electron, and Ae is acceleration of the electron)

Fp = Mp × Ap (Mp is mass of the proton, and Ap is acceleration of the proton)

⇒Me × Ae = Mp × Ap

⇒ Ae = Mp × Ap / Me

6) Now, state the equations for the velocity in uniformly accelerated motion:

i) Vf² = Vo² + 2ad

Vo² = 0 for both cases, and d is the same distance.

⇒ Vf² = 2ad

ii) For the proton Vf² = 2(Ap)(d) ⇒ Ap = Vf² / (2d)

⇒ Ap = (55,000 m/s)² / (2d)

iii) For the electron Vf² = 2(Ae)² (2d)

iv) Using Ae = Mp × Ap / Me (found prevously):

Vf² = Mp × (55,000 m/s)² / (2d) × (2d) / Me

⇒ Vf² = Mp × (55,000 m/s)² / Me

Taking square root in both sides:

⇒ Vf = 55,000 m/s × √ [Mp / Me]

7) These are the values for the masses of a proton and an electron:

Mp = 1.67 × 10⁻²⁷ kg

Me = 9.11×10⁻³¹ kg

8) Replace and compute:

Vf = 55,000 m/s × √ [ 1.67 × 10⁻²⁷ kg / 9.11×10⁻³¹ kg] = 2,354,841.8 m/s

Round to two significan digits: Vf = 2,400,000 m/s

SVEN [57.7K]3 years ago
3 0
The correct answer to this question is this one:

You will need these relations

<span>qV = KE = 1/2m<span>v^2

</span></span>

You know the charge of a proton (the elementary charge in this case), you know its mass, and you know it's velocity. From there you can find its kinetic energy and from there you can find it's voltage. Once you know the potential difference there all you have to do is reverse the process. Multiply the charge of an electron by V and then solve for the velocity using the mass of the electron

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A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is 1.37 m/s^2

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

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- The frictional force, F_f, backward

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v^2-u^2=2as

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v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

a=1.37 m/s^2

s = 350 m

Solving for v,

v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s

4)

For this part of the problem, we can use the following suvat equation:

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Here we have:

v = 5.2 m/s

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