```1.  Use Coulomb's Law for the E field.  For the field
to be zero, the point must be sqrt2 farther from the
+2 C charge and at a position where the two contributions
to the field are in opposite DIRs.  Only Region III is
possible.

2.  3 is greatest because both distances are one unit
and the two field contributions have same DIR.
1 is least because both distances are one unit
and the two field contributions have opposite DIRs.
4 and 5 clearly tie.
2 is nonzero, but less than 4 or 5 because the
two field contributions are in opposite DIRs.

3.  Let kq/d^2 represent the field due to one unit of
charge which is 1 unit away.  Numbers below are in
terms of this unit of field.
2 is 1 unit charge 2 units away= 0.25.
1 is (1+.25+1/9)=1.36
4 is (2+.25)
3 is (2-.25)=1.75
5 is (1-.25-1/9)=0.64

4.  The charge must be POS and to left since the
field gets smaller to the right.  The distance
squared is inversely proportional to the field
strength, thus distances squared should be 4/9,
thus distances are 2/3.

5.  In all cases, charges are shared equally, thus
only total charge matters, with POS considered
larger than NEG for this question only.

6.  Find the total E field, then mult by 30 muC.
Total field is k(40muC)((1/2m)^2-(1/3m)^2),
where the minus appears because the field due
to the NEG charge at 3 m is in the opposite
DIR of the larger contribution.

7.  Add two vector fields, then mult. by 32 nC.
Ex=k((60nC/(.005m)^2)+(80nC/(.005m)^2))cos(37degrees)
Ey=k(-(60nC/(.005m)^2)+(80nC/(.005m)^2))sin(37degrees)
E=sqrt(Ex^2+Ey^2)
F=qE=1.30 N

Ex=k(16nC)((1/4m)^2+(1/5m)^2*cos(37)+0)
Ey=K(16nC)((1/3m)^2+0+(1/5m)^2*sin(37))
E=sqrt(Ex^2+Ey^2)=23.7 N/C

9.  Integrate for the magnitude of x and y components:
with lambda=25nC/(0.5piR) and R=2.0m
|Ex|=|(k(lambda)Rdtheta/R^2)*cos(theta)) integrated
from zero to 90 degrees|=k(lambda)/R
|Ey|=|(k(lambda)Rdtheta/R^2)*sin(theta)) integrated
from zero to 90 degrees|=k(lambda)/R
E=sqrt(Ex^2+Ey^2)=sqrt2*k(lambda)/R=50.6 N/C

10.  Use Gauss' Law.  The dimensions of the box
and the charges outside the box make no difference.
Flux=total charge inside/epsilon0
= -5nC/epsilon0=-565 Nm^2/C

11.i. Use Gauss' Law and the fact that the
field in the metal is zero.=> -Q
ii. Use i. and the fact that the metal is
neutral=>+Q
iii. Both are conducting. All the charge
moves to the outer surface of the shell.=> 0
iv. 0
v. +Q

12.i. Nowhere is the field zero.  (NOTE: there
is a bug in the electronic version of this
question, which accepts 6 as the correct
ii. The 4 faces for which the field
runs along the face=>EdotdA=0.

13.  Use Gauss' Law.  Draw a GSurface
at radius r outside the sphere.
Qenc = (4/3)pi(.12m)^3*3nC/m^3 = 217 pC.
In terms of r, the flux = 4pir^2E.  So
E=kQenc/r^2=86.9 N/C

14. Ignore the outer surface by Gauss' Law.
E=k4pi(0.01m)^2*(40 pC/m^2)/(0.02m)^2=1.13 N/C.

15. Let kq/d represent the PTL due to one
unit of charge and distance.  In those units,
the PTL at F4 is (1+(3/3))=2, i.e. +2kq/d.
For PTL at F4=0 we need to add -2kq/d; since
the charge is -2q the distance is 1d, so
points F3, G4, and F5 would work.

16. The PTL is lowest at the collection of
NEG charge and highest at the collection of
POS charge.  The plates are to be considered
infinite, thus 1 and 2 tie, and 4 and 5 tie.

17. Let c=sqrt(a^2+b^2)=5 m.
VA=k((q/a)+(Q/c))
VB=k((Q/a)+(q/c))
Since Q=-q
VA=kq((1/a)-(1/c))
VB=kq(-(1/a)+(1/c))
So VA-VB=2kq((1/a)-(1/c))=4.8 V

18. The deltaPTL is minus the integral of Edotdr.
Since the Efield is 12 N/C in the +x DIR,  the
dot product eliminates the y information, and
the integral is just -E times x evaluated at
upper limit 3 m and lower limit 8 m.
It's easier just to draw the picture with the
constant E pointing towards lower PTL.
So VA-VB=-(12 N/C)*(3 m - 8 m)=+60 N*m/C=+60 V.
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