Calculate Electric Field & Energy in Capacitors: Physics Problems #1-3

[fish]

problem #1
Two point charges of +3.0uC and -7.0uC are placed at x=0 and x=.20m, respectively. What is the magnitude of the electric field at the point midway between them (x=.10m)?

E3 = (kq)/r^2 = [(9x10^9 )(3x10^-6 C)]/.10m^2 = 2.7x10^6 N/C
E7 = (kq)/r^2 = [(9x10^9 )(7x10^-6 C)]/.10m^2 = 6.3x10^6 N/C
E=E3+E7 = 9x10^6 N/C
solutions manual has the answer as 9x10^5 N/C


problem #2
If the distance between two closely spaced parallel planes is halved, the electric field between them: becomes 4 times as large.

E=4/A
E=4/d^2
if d=2, E=1
if d=1, E=4
so becomes 4 times as large
solutions manual has the answer as remains the same


problem #3
A 12-V battery is connected to a parallel-plate capacitor with a plate area of 0.40 m^2 and a plate separation of 2.0 mm. How much energy is stored in the capacitor?

A=0.40 m^2
d=2.0 mm = .002 m
V= 12V
find Uc

C=(EoA)/d = (8.85x10^-12 x .40 m^2)/.002 m
C= 1.77x10^-9 F
Uc=1/2cv^2 = 1/2(1.77x10^-9 F)(12v)^2 = 1.27x10^-7 J

solutions manual has the answer as 2.5x10^-7 J

 

[Doc Al]

Your answers to #1 and #3 look good to me. I'm not sure what you are doing for #2, but in any case: as long as the charge remains the same, the field within the (closely spaced) plates is independent of separation.

 

[fish]

thanks Doc, in #2 I was getting confused about the (A)area= d^2 vs the separation distance. Q remains same, E doesn't depend on separation distance between two closely spaced parallel plates.