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Discussion Question 4DP212, Week 4

Superposition and Conductors

As we saw in the last problem, superposition is an extremely useful tool in potential problems.However, we have to be very careful with superposition when conductors are around letsexplore this!

Two thin plates of infinite area and made of insulating material are on either side of the originand a distance d=5 cm away from it. They carry uniformly-distributed surface charges withthe values given below. In this problem, we will be concerned with the electric potentialdifference between the two plates. For convenience, well use the left-hand plate as ourreference and set the electric potential to zero there. (This is indicated by the ground symbol in

the figure.)

(a) What is the electric potential Vb of the right-hand plate?

E=

a

b

2o

xis constant between the plates.

Vb Va = Exdxd

d

Vb 0 =

a

b

2o

dxd

d

=

a

b

2o

2d= 3.39 1010 V

(b) Now suppose a positive point charge Q = +5 C is placed at the location

(x,y) = (-5 cm, +2 cm). What is the electric potential difference Vbetween the points(x,y) = (d,0) and (-d,0) on the x axis? (Remember, superposition )

Superposition: Add the effect of the point charge to that of the two plates:Due to Q:

VbV

a=

kQ

(d+ 5 cm)2 + 2 cm( )2

kQ

(d+ 5 cm)2 + 2 cm( )2

=kQ

(2d)2 + 2 cm( )2

kQ

2= 2.111012 J/C

total VbV

a=

a

b

o

d+kQ

(2d)2 + 22kQ

2= 2.07 1012 J/C

x

ba ya = -4 C/m

2

b = +2 C/m2

d= 5 cm

d d

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Now get rid of the point charge. Instead,suppose that an uncharged metal slab ofthickness 2 cm is placed parallel to the twoplates and centered on the origin.

Does superposition still work?YES we just have to be a bit careful.

(c) If we considered the uncharged metal slab all on its own, in

isolation, what potential difference Vb Va would it cause betweenthe two plates?0 (produces no fields)

(d) If we considered the charged insulating sheets all on theirown, in isolation, what potential difference Vb Va would theycause between the two plates?Same as in part (a)

Pure superposition would suggest that adding these twocontributions together gives the correct net result. But not quite

(e) Calculate thefullpotential difference Vb Va between the plates: find the electric fieldeverywhere between the plates, and integrate it to find Vb Va.

Emust = 0 inside the metal slab. The external field induces charge densities L and Ron thetwo surfaces of the slab, and these charges contribute to the potential difference, Vb Va. You

could solve for L and Rusing the techniques you learned last week, but there is a simpler way,using things you already know:

L = R, because the slab is electrically neutral.Therefore, outside the slab they cancel and do not affect the electric field outside the slab.

E= 0 inside the slab.So, Vb Va = the same integral before, except thatEx is only nonzero over a distance 2dw.

VbV

a=a b

2

2dw( ) = 2.711010

V

(f) Can you explain why adding the results of (c) and (d) together did not work?Conductors respond to the presence of charged objects by internally rearrangingtheir owncharges. Thus, when we apply superposition to a collection of objects that includes one or moreconductors, we must remember that the conductors are affected by the other objects. If you wantto use superposition, you must use thefinalcharge distribution.

Vb

d d

Va

Vb

d d

Va

ba

d d x

unchargedmetal slab

w = 2 cm

y