Conductors

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Discussion

introduction

Static charge moves to the outside. This sounds stupid, "static charge moves".

Field is zero on the inside. Faraday cage, electrostatic shielding, etc.

Field perpendicular to surface.

Charge collects at points. This leads to a phenomena known as corona discharge. "Corona" from the latin word for crown.

Corona Discharge from a Wire

Corona discharge is a useful way to passively eliminate charge.

Lightning Rods …
on an Airplane Wing on an Electric Power Substation

text

Summary

Problems

practice

  1. Two metal spheres are electrically charged. One of them carries +6 μC and the other −12 μC. The two spheres are carefully brought in contact and then separated. What is the new charge on each sphere if …
    1. the spheres are the same size
    2. the positive sphere has twice the radius of the negative sphere

    Solutions …

    1. With two spheres of equal size, the total charge will try to distribute itself evenly between them.The positive charges are repelled by one another and try to get as far apart as possible. The negative charges behave the same way. They best strategy for maximizing separation is to send half your members to one sphere and half to the other. Since neither sphere is more "attractive" than the other, the split should be an even one. On the whole, we have …
       
      Qnet = (+6 μC) + (−12 μC) = −6 μC
       
      of charge to slit evenly into …
           
       Q1 = Q2 =  −6 μC  = −3 μC  
      2
           
      for each sphere.
    2. If you were a charge, running away from the other charges and you had to decide between two conductors of different size, which would you choose? Well, you can't all choose the bigger one because then you'd be closer together than if some of you chose the big one and some the small one. Would you divide up evenly? No, that doesn't make much sense either. The small sphere would be pack tight and the big one would have unused capacity for holding charge. No, the "logical" conclusion in the minds of our anthropomorphic charges is to distribute themselves at arm's length against the outside of each sphere. The next thing to do is test the common wisdom with mathematics.

      The quantity that determines the motion of charges is voltage -- potential difference, that is. Charge will flow from high potential to low potential in much the same way that water flows from regions of high altitude to regions of low altitude. Our two charged spheres are like two lakes. Putting the spheres in contact is like connecting two lakes with an aqueduct. The higher one (higher in altitude) will drain away into the lower one until both lakes are at the same level. Similarly, our two charged spheres will exchange charge until they both reach the same voltage.
                       
      V1 = V2 ⇒ kq1  =  kq2 ⇒ q1  =  q2
      r1 r2 2r r
                       
      At the same time, charge must be conserved. We started with −6 μC overall and so we shall end. Mathematically, we write this as …
           
      q1 + q2 = −6 μC ⇒ q2 = −6 μC − q1
           
      All that remains is to combine the two equations and solve for the two unknowns.
           
      q1  =  −6 μC − q1
      2r r
           
      q1  =  −12 μC − 2q1
      3q1  =  −12 μC
           
      q1  =  −4 μC
      q2  =  −2 μC
           
      Well whad'ya know. The one that's twice as big (has twice the radius) can hold twice as much charge. That makes sense.
  2. Write something.
    • Answer it.
  3. Two conducting spheres are concentrically nested as shown in the cross-sectional diagram below. The inner sphere has a diameter of 3 cm and a net charge of +12 μC. The outer spherical shell has an inner diameter of 6 cm, an outer diameter of 8 cm, and a net charge of −6 μC.
     
     
    1. Determine the net charge on the …
      1. inner surface of the outer spherical shell
      2. outer surface of the outer spherical shell
    2. Complete the following table.

      distance from
      center (cm)      		 electric field (MV/m) electric potential (MV)
      direction magnitude sign magnitude
      0        
      1        
      2        
      3        
      4        
      5        
      6        
      7        
      8        
      9        
      10        

    3. Sketch the following quantities as functions of distance from 0 to 10 cm.
      1. magnitude of the electric field
      2. electric potential
    Solution …
    1. The charge will distribute itself so as to "screen out" the electric field within the hollow sphere. There will be just as much charge on the inside of the surface as there is on the nested sphere with the total charge adding up to −6 μC. Thus there must be …
      1. −12 μC of charge on the inner surface and …
      2. +6 μC of charge on the outer surface.
    2. Now that we know the charge distribution we can determine the electric field by repeated application of Gauss' law. Imagine a spherical Gaussian surface concentric with the nested spheres and watch its radius vary from 0 to 10 cm.
       
      ∯ E · dA =  Q  ⇒  E (4πr2) =  Q  ⇒  E =  1   Q  =  kQ
      ε0 ε0 4πε0 r2 r2
       
      For the first 3 cm the Gaussian sphere contains no charge, which means there is no electric field. That's the way it works in a conductor. There can be no field inside a conductor once the charges find their equilibrium distribution.
                   
       
      E =  kQ
      r2
      		 = 
      (9.0 × 109 N m2/C2)(0 C)
      r[m]2
      		 =  0 V/m  (0 cm to 3 cm)
       
      When the radius reaches 3 cm the Gaussian sphere finally contains some charge
       
      Qnet = +12 μC
       
      No additional charge makes it into the Gaussian sphere for awhile.
       
       
      E =  kQ
      r2
      		 = 
      (9.0 × 109 N m2/C2)(+12 × 10−6 C)
      r[m]2
      		 = 
      +108,000 Vm
      r[m]2
      		 (3 cm to 6 cm)
       
      When the radius reaches 6 cm, the Gaussian surface contains no net charge again.
       
      Qnet = (+12 μC) + (−12 μC) = 0 μC
       
      No charge, no field. (Or is it the other way round?)
       
       
      E =  kQ
      r2
      		 = 
      (9.0 × 109 N m2/C2)(0 C)
      r[m]2
      		 =  0 V/m  (6 cm to 8 cm)
       
      When the Gaussian surface makes it to the outside of the outer sphere, the net charge is nonzero once again. This time it's
       
      Qnet = (+12 μC) + (−6 μC) = +6 μC
       
      Thus …
       
       
      E =  kQ
      r2
      		 = 
      (9.0 × 109 N m2/C2)(+6 × 10−6 C)
      r[m]2
      		 = 
      +54,000 Vm
      r[m]2
      		 (8 cm to ∞)
       
      Numbers in. Answers out. See the table below for the summary of all this talk. By the way, the positive signs in the answers tell us that the field is directed radially outward in the places where it exists.
       
      Now, on to the potential. In this problem, it's best to compute electric potential from the electric field using the mutated form of the work-energy theorem. Imagine a small, positive test charge. Drag it from infinity to any point a distance r from the center of the spheres. The work per unit charge is the electric potential and the force per unit charge is the field.
        r     r
      V = −  ⌠
      ⌡      		 E ·dr  ⇐  U = −  ⌠
      ⌡      		 F ·dr
        ∞     ∞
      Apply this formula in a piecewise manner, noting the discontinuities in the electric field at the surfaces of the conductors. Work from the "outside in" as is the custom with potentials.
       
      For the space surrounding the outer sphere we get …
       
        r  
      V = −  ⌠
      ⌡      		 kQ  dr
      r2
        ∞  
      		 = 
      (9.0 × 109 N m2/C2)(+6 × 10−6 C)
      r[m]
      		 = 
      +54,000 Vm
      r[m]
      		 (8 cm to ∞)
      Electric potential in a conductor is the same everywhere, is the same as the potential at the last place it was computable -- the surface.
       
        0.08 m
      V = −  ⌠
      ⌡      		 kQ  dr
      r2
        ∞
      		 = 
      (9.0 × 109 N m2/C2)(+6 × 10−6 C)
      0.08 m
      		 =  0.675 MV  (6 cm to 8 cm)
      Now, in the gap between the conductors we see the first application of piecewise integration.
        0.08 m   0.06 m   r
      V = −  ⌠
      ⌡      		 E ·dr  −  ⌠
      ⌡      		 E ·dr  −  ⌠
      ⌡      		 E ·dr
        ∞   0.08 m   0.06 m
      V = +  0.675 MV  +  0 MV  +  kQ  ⎡
      ⎣      		 1  −  1 ⎤
      ⎦
      r[m] 0.06 m
       
      V = + 0.675 MV + 0.108 MVm  ⎡
      ⎣      		 1  −  1 ⎤
      ⎦      		  (3 cm to 6 cm)
      r[m] 0.06 m
       
      My, that was unpleasant.

      We finish by evaluating this last expression at 3 cm. That's the potential from the surface of the inner sphere all the way to the center.
       
      V = + 0.675 MV + 0.108 MVm  ⎡
      ⎣      		 1  −  1 ⎤
      ⎦      		  = 2.475 MV (0 cm to 3 cm)
      0.03 m 0.06 m
       
      Once again, check the table below for the results of all this work.

      distance from
      center (cm)      		 electric field (MV/m) electric potential (MV)
      direction magnitude sign magnitude
      0 n/a 0 + 2.475
      1 n/a 0 + 2.475
      2 n/a 0 + 2.475
      3 n/a, out 0, 120 + 2.475
      4 out 67.5 + 1.575
      5 out 43.2 + 1.035
      6 out, n/a 30.0, 0 + 0.675
      7 n/a 0 + 0.675
      8 n/a, out 0, 8.44 + 0.675
      9 out 6.67 + 0.600
      10 out 5.40 + 0.540


    3. And lastly, graphs of …
      1. magnitude of the electric field
      2. electric potential
  4. Write something completely different.
    • Answer it.

conceptual

  1. Read the following passages from Benjamin Franklin.
     
    In September 1752, I erected an Iron Rod to draw the Lightning down into my House, in order to make some Experiments on it, with two Bells to give Notice when the Rod should be electrify'd. A contrivance obvious to every Electrician.
     
    In Philadelphia I had such a rod fixed to the top of my chimney, and extending about nine feet above it. From the foot of this rod, a wire (the thickness of a goose-quill) came through a covered glass tube in the roof, and down through the well of the staircase; the lower end connected with the iron spear of a pump. On the staircase opposite too my chamber door, the wire was divided; the ends separated about six inches, a little bell on each end; and between the bells a little brass ball, suspended by a silk thread, to play between and strike the bells when clouds passed with electricity in them.
    [extend these quotes]
     
    Here are a few portraits of Franklin seated besides his device. If you look very closely [magnify] you can see the "little brass ball suspended by a silk thread" between the bells.
           
       
        Sources: Unknown  
    1. When a storm cloud passes overhead, the bells become charged opposite each other. Explain why this happens
    2. When the bells become charged, the brass ball will bounce back and forth between the two bells. Explain why this happens.
    3. What could this device be used for?
    4. Why don't you have one in your home today?

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