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Two parallel plate capacitors of capacity \mathrm{C}$ and $3 \mathrm{C} are connected in parallel combination and charged to a potential difference 18 \mathrm{~V}. The battery is then disconnected and the space between the plates of the capacitor of capacity \mathrm{C} is completely filled with a material of dielectric constant 9 . The final potential difference across the combination of capacitors will be___________\mathrm{V}.

Option: 1

6


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Initially, both the capacitor are changed such that the potential drop across each capacitor is \mathrm{18 \mathrm{~V}}
Affer the insertion of dielectric, the capacitance of first capaitor (c) changes to KC
\mathrm{V_0 \rightarrow} common potential at steady state
By charge conservation,

\mathrm{q_1+q_2=q_1'+q_2' }

\mathrm{C(18)+3 C(18)=k C\left(v_0\right)+3 C\left(v_0\right) }

\mathrm{18+54=9 v_0+3 v_0 }
\mathrm{v_0=6 \mathrm{~V}}
The final potential difference across the combination of capacitors will be \mathrm{6 \mathrm{~V}}

Posted by

Gaurav

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