Gauss's Law

Practice

practice problem 1

The electric field near the surface of the earth has a magnitude of approximately 150 V/m and points downward. Determine the following quantities for the earth …
  1. the net charge (including sign)
  2. the surface charge density in coulombs per square meter (including sign)
  3. the surface charge density in elementary charges per square meter (indicate wheter the charge is due to an excess or deficit of electrons)

solution

  1. Answer it.
  2. Answer it.
  3. Answer it.

practice problem 2

Two conducting spheres are concentrically nested as shown in the cross-sectional diagram to the right. The inner sphere has a radius of 3 cm and a net charge of +12 μC. The outer spherical shell has an inner radius of 6 cm, an outer radius 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

practice problem 3

Write something.

solution

Answer it.

practice problem 4

Write something completely different.

solution

Answer it.