JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
Two parallel plate capacitors of capacity \(C\) and \(3\,C\) are connected in parallel combination and charged to a potential difference \(18\,V\). The battery is then disconnected and the space between the plates of the capacitor of capacity \(C\) is completely filled with a material of dielectric constant \(9\). The final potential difference across the combination of capacitors will be \(V\)
- A \(5\)
- B \(4\)
- C \(6\)
- D \(1\)
Answer & Solution
Correct Answer
(C) \(6\)
Step-by-step Solution
Detailed explanation
Initial charge on \(C =18\,CV\) initial charge on \(3 C =54\,CV\) Let final common potential difference \(= V ^{\prime}\) \(9\,CV ^{\prime}+3\,CV ^{\prime}=18\,CV +54\,CV\) \(12\,CV ^{\prime}=72\,CV \Rightarrow V ^{\prime}=6 V\)
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