Calculating the Effects of Removing Dielectric from Parallel Plate Capacitor

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wrthwrld
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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
[tex]\kappa[/tex]= E[tex]_{}[/tex]/E

I don't get what this is asking for or how to carry it out
 
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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 [tex]k = \frac {\epsilon} { \epsilon_0}[/tex]