Answer:
Applied voltage should be 13.5396 kV
Explanation:
The charge stored by a capacitor when subjected to a potential difference 'v' across it's plates is given by
[tex]Q=\frac{K\epsilon _{o} A}{d}V[/tex]
Solving for V we get
[tex]V=\frac{Qd}{K\epsilon _{o}A}[/tex]
thus to store a charge of 0.11mC we solve for V as follows
Applying values we get
[tex]V=\frac{0.11\times 10^{-3}\times 0.01\times 10^{-3}}{27\times 10^{-4}\times 3.4\times 8.85\times 10^{-12}}[/tex]
[tex]\therefore V=13.5396kV[/tex]