Problem on two parallel plate capacitor.

[alexander3133]

An air-filled parallel-plate capacitor with plate separation d and plate area A is connected to a battery which supplies a voltage Vo between plates. With the battery left connected, the plates are moved apart to a distance of 10d. Determine by what factor each of the following quantities changes:

a) Vo
b) capacitance, C
c) electric field, E
d) electric flux density, D
e) charge, Q
f) charge density
g) energy stored, U

An explaination on each answer will be highly appreciate, thanks.

 

[walid1]

I think but not sure the following:
You know that C = e0 (A / d)
where A is the area of the capacitor plates, and d is the distance between them.

e0 is the permittivity of free space (8.85X10-12)

a) Vo is still as lt is cause the relation did not affected by batt voltage. the capacitance of a capacitor will is a fix value when it is manufactured>

b) capacitance, C: as d increase C decrease by the same factor (10 times) cause the relation is inversly proporetional
  the following link ia a calculator to see by yourself
http://micro.magnet.fsu.edu/electromag/java/capacitance/

c) electric field, E: V = Ed = (F/q)*d = Fd/q= W/q,  since q (the charge) is at the demonator then E increased 10 times

d) electric flux density, D: I don't know may be later I can found somthing
e) charge, Q: from the relation (Q = CV ) we see Q is directly proportinal to C, so Q is decreased 10 times as C
f) charge density: is decreased 10 times as Q
g) energy stored, U: W =