Plate Capacitor

1. Feb 16, 2010

wrthwrld

A parallel plate capacitor of capacitance 10microfarads has the space between filled with material with
(dielectric constant k)= 4.0. The capacitor is charged to a potential difference of 2.0V . With the capacitor connected
to the battery the dielectric is removed.
1. The capacitance is F.
2. The potential difference across the capacitor is V .
3. The charge on the plates is C.
4. The energy stored in the capacitor is J.

F=C/V
$$\kappa$$= E$$_{}$$/E

I dont get what this is asking for or how to carry it out

2. Feb 16, 2010

willem2

I don't see a question. You seem to have included some weird latex codes that only produce a
large grey square while trying to write $$k = \frac {\epsilon} { \epsilon_0}$$