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class code: SPS22 teacher: Mr. Elert
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email: elert@midwoodscience.org

Physics A: Problem Set 5: Electric Potential

recommended reading

Barron's Let's Review: 8.10–8.11 Potential Difference
physics.info: Electric Potential
Wikipedia: Electric potential
HyperPhysics: Voltage
Khan Academy: Electric potential energy, Voltage, Volt

classwork

  1. A typical television picture tube (also known as a cathode ray tube or CRT) is about a half meter long and has a potential difference of about 10,000 V between the front and the back. Determine the following quantities for an electron in a typical CRT when it strikes the phosphor coating on the front of the tube.
    1. its kinetic energy
    2. its velocity
    Determine the following quantities inside a typical CRT TV.
    1. the acceleration of an electron
    2. the force on an electron
    3. the electric field strength
    Electrons with this much energy will produce x-rays when stopped abruptly.
    1. Are the x-rays produced by a CRT television a matter for concern? If yes, why? If not, why not?

    1. The electron gains kinetic energy because work was done on it. Work is a part of the definition of voltage. Solve for work and you've effectively solved for kinetic energy. If you don't know the charge on the electron, look it up.

      V = 
      W  =  K
      q q
      K =  qV
      K =  (1.60 × 10−19 C)(10,000 V)
      K =  1.60 × 10−15 J
    2. Do you remember the defintition of kinetic energy? Use that equation to get the speed of the electron. If you don't know the mass of the electron, look it up.

      K =  ½mv2
      v =  √(2K/m)
      v =  √[2(1.60 × 10−15 J)/(9.11 × 10−31 kg)]
      v =  5.93 × 107 m/s
    3. Acceleration is found with one of those equations of motion — the one that doesn't include time.

      v2 = v02 + 2as

      Solve for acceleration and remember that the electron started from rest.

      a =  v2/2s
      a =  (5.93 × 107 m/s)2/(2 × 0.5 m)
      a =  3.51 × 1015 m/s2
    4. You've got acceleration. You've got mass. You multiply them to get force (because, you know, Newton's second law and all that).

      F =  ma
      F =  (9.11 × 10−31 kg)(3.51 × 1016 m/s2)
      F =  3.20 × 1015 N

      An alternate method is to use the definition of work and solve it for force. Recall that the work done on an electron in the CRT is the same as its kinetic energy when it strikes the screen.

      W =  Fd 
       
      F = 
      W  =  K
      d d
       
      F =  1.60 × 10−15 J  
      0.50 m  
      F =  3.20 × 1015 N  
       
    5. Field is force per charge (because, you know, that's what it is).

      E = 
      F
      q
       
       
      E =  3.20 × 1015 N  
      1.60 × 10−19 C  
      E =  20,000 N/C  
       
    6. The glass in the screen of a CRT contains lead, which absorbs x-rays. You do not need to be concerned about the the x-rays produced by a CRT television (especially if you don't have one).

  2. The inside of a human nerve cell is more negative than the outside by about −70 mV. When a nerve impulse propagates down an axon, the polarity reverses and the inside is more positive than the outside by +40 mV. This action potential lasts 2 ms and then the original resting potential is restored. All of this takes place in the space of about 6 nm, the thickness of the cell membrane.
    1. What is the magnitude and direction of the electric field (in V/m) across the membrane of a neuron during…
      1. the resting phase
      2. the action phase
    2. How much work is done moving a single sodium ion (Na+) across the cell membrane of a neuron? State your answer in…
      1. joules
      2. electron volts
    3. What is the power of this microscopic event?

    Nerve cell membrane showing ion flow, surface charge, and voltage during resting and action potentials

    1. Let's start with direction, since that doesn't require a calculator. Electric field lines start on positive charges and end on negative charges. The electric field points in to the cell during the resting potential when the interior is negative and out of the cell during the action potential when the interior is positive.

      Compute the field strength using this handy equation. Do it once for the resting potential.

      E = 
      V
      d
      E =  70 × 10−3 V
      6 × 10−9 m

      E = 1.17 × 107 V/m in

      Do it again for the action potential.

      E = 
      V
      d
      E =  40 × 10−3 V
      6 × 10−9 m

      E = 6.67 × 106 V/m out

    2. Take the definition of voltage.

      ΔV =  W
      q

      solve it for work.

      W = qV

      Numbers in. Answer out. The charge on a sodium ion is one elementary charge. Use the potential difference between the action potential and the resting potential. We'll do this first using pure SI units.

      W = (1.6 × 10−19 C)[(+40 × 10−3 V) − (−70 × 10−3 V)]

      W = 1.76 × 10−17 J

      Now we'll do it again with acceptable, non-SI units.

      W = (1 e)[(+40 mV − (−70 mV)]

      W = 110 meV or 0.110 eV

    3. Power is the rate at which work is done. Use SI units for this part.

      P = 
      W
      t
      P =  1.76 × 10−20 J
      2 × 10−3 s

      P = 8.80 × 10−15 W

homework

  1. In a region where the electric field is constant, as it is between two oppositely charged parallel plates, is the voltage also constant? Explain your answer.

    In order for there to be an electric field, there has to be an electric potential difference, so no — the voltage is not constant in an area where the field is constant. In the case of a parallel plate capacitor (a of charge storage device found in many electrical devices), one of the plates would be at a positive voltage and the other at a negative voltage. The voltage difference is what makes the field and the parallel geometry of the plates is what makes the field uniform.

  2. Two related questions.
    1. What do the electric fields lines look like in a region where the magnitude of the electric field is uniform?
    2. What do they look like in a region where the electric potential is uniform?
    1. The strength of the electric field is shown in diagrams by the spacing between the field lines. The closer together the field lines are, the stronger the field is. The farther apart they are, the weaker the field is. In a region where the field strength is constant, the spacing between the field lines is also constant. In a region where the magnitude of the electric field is uniform the field is drawn as a series of uniformly spaced, parallel lines.
    2. Electric field lines are driven by a difference in electric potential in much the same way that streamlines in a river are driven by a difference in height. In a region where there is no difference in height (like a puddle), there is no flow of water. In a region where the electric potential is uniform (like a Faraday cage), there is no electric field.