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Electrical potential energy converted to kinetic energy

Calculate the final speed of a free electron accelerated from rest through a potential difference of 100 V. (Assume that this numerical value is accurate to three significant figures.)

Strategy

We have a system with only conservative forces. Assuming the electron is accelerated in a vacuum, and neglecting the gravitational force (we will check on this assumption later), all of the electrical potential energy is converted into kinetic energy. We can identify the initial and final forms of energy to be KE i = 0, KE f = ½ mv 2 , PE i = qV , and PE f = 0.

Solution

Conservation of energy states that

KE i + PE i = KE f + PE f .

Entering the forms identified above, we obtain

qV = mv 2 2 . size 12{ ital "qV"= { size 8{1} } wideslash { size 8{2} } ital "mv" rSup { size 8{2} } "." } {}

We solve this for v size 12{v} {} :

v = 2 qV m . size 12{v= sqrt { { {2 ital "qV"} over {m} } } "." } {}

Entering values for q , V , and m size 12{q, V", and "m} {} gives

v = 2 –1.60 × 10 –19 C –100 J/C 9.11 × 10 –31 kg = 5.93 × 10 6 m/s.

Discussion

Note that both the charge and the initial voltage are negative, as in [link] . From the discussions in Electric Charge and Electric Field , we know that electrostatic forces on small particles are generally very large compared with the gravitational force. The large final speed confirms that the gravitational force is indeed negligible here. The large speed also indicates how easy it is to accelerate electrons with small voltages because of their very small mass. Voltages much higher than the 100 V in this problem are typically used in electron guns. Those higher voltages produce electron speeds so great that relativistic effects must be taken into account. That is why a low voltage is considered (accurately) in this example.

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Section summary

  • Electric potential is potential energy per unit charge.
  • The potential difference between points A and B, V B V A size 12{V rSub { size 8{B} } -V rSub { size 8{A} } } {} , defined to be the change in potential energy of a charge q size 12{q} {} moved from A to B, is equal to the change in potential energy divided by the charge, Potential difference is commonly called voltage, represented by the symbol Δ V size 12{V= { {"PE"} over {q} } "." } {} .
    Δ V = ΔPE q and ΔPE = q Δ V . size 12{?V= { {?"PE"} over {q} } " and "D"PE="q?V "." } {}
  • An electron volt is the energy given to a fundamental charge accelerated through a potential difference of 1 V. In equation form,
    1 eV = 1.60 × 10 –19 C 1 V = 1.60 × 10 –19 C 1 J/C = 1.60 × 10 –19 J.
  • Mechanical energy is the sum of the kinetic energy and potential energy of a system, that is, KE + PE . size 12{"KE"+"PE"} {} This sum is a constant.

Conceptual questions

Voltage is the common word for potential difference. Which term is more descriptive, voltage or potential difference?

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If the voltage between two points is zero, can a test charge be moved between them with zero net work being done? Can this necessarily be done without exerting a force? Explain.

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What is the relationship between voltage and energy? More precisely, what is the relationship between potential difference and electric potential energy?

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Voltages are always measured between two points. Why?

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How are units of volts and electron volts related? How do they differ?

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Problems&Exercises

Find the ratio of speeds of an electron and a negative hydrogen ion (one having an extra electron) accelerated through the same voltage, assuming non-relativistic final speeds. Take the mass of the hydrogen ion to be 1 . 67 × 10 27 kg . size 12{1 "." "67"×"10" rSup { size 8{-"27"} } " kg" "." } {}

42.8

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Questions & Answers

What is conductivity
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PhysicswithMrV
d=1.0km÷1000=0.001 t=5×60=300s s=d\t s=0.001/300=0.0000033m\s
Serah
A puck is moving on an air hockey table. Relative to an x, y coordinate system at time t 0 s, the x components of the puck’s ini￾tial velocity and acceleration are v0x 1.0 m/s and ax 2.0 m/s2 . The y components of the puck’s initial velocity and acceleration are v0y 2.0 m/s and ay 2.0
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becuse the d .c cannot travel for long distance trnsmission
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Practice Key Terms 4

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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