Question:
Two condensers of capacities, '2C' and 'C' are joined in parallel and charged upto potential 'V'. The battery is then disconnected and condenser of capacity 'C' is filled completely with a medium of dielectric constant 'K'. The potential difference across the capacitors in the second case is
Two condensers of capacities, '2C' and 'C' are joined in parallel and charged upto potential 'V'. The battery is then disconnected and condenser of capacity 'C' is filled completely with a medium of dielectric constant 'K'. The potential difference across the capacitors in the second case is
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The battery is disconnected, so total charge stays constant.
Updated On: Oct 1, 2026
- \(\frac{3\text{V}}{(\text{K}+2)}\)
- \(\frac{\text{V}}{(\text{K}+2)}\)
- \(\frac{5\text{V}}{(\text{K}+2)}\)
- \(\frac{2\text{V}}{(\text{K}+2)}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Two capacitors \(2C\) and \(C\) in parallel have a total capacitance \(3C\). Charged to \(V\), the total charge is \(Q = 3CV\). After disconnecting the battery, \(Q\) cannot change.
Step 2: New capacitance:
Filling \(C\) with a dielectric of constant \(K\) makes it \(KC\). The parallel combination now has \(2C + KC = (K+2)C\).
Step 3: New potential:
\[ V' = \frac{Q}{C_{new}} = \frac{3CV}{(K+2)C} = \frac{3V}{K+2} \]
Step 4: Check:
Option (A). For \(K = 1\) (no dielectric), \(V' = V\), as expected. Options (B), (C), (D) do not give \(V\) at \(K = 1\).
Final Answer:
Total charge is fixed, capacitance becomes (K + 2)C. \[ \boxed{\text{(A) }\dfrac{3V}{K+2}} \]
Two capacitors \(2C\) and \(C\) in parallel have a total capacitance \(3C\). Charged to \(V\), the total charge is \(Q = 3CV\). After disconnecting the battery, \(Q\) cannot change.
Step 2: New capacitance:
Filling \(C\) with a dielectric of constant \(K\) makes it \(KC\). The parallel combination now has \(2C + KC = (K+2)C\).
Step 3: New potential:
\[ V' = \frac{Q}{C_{new}} = \frac{3CV}{(K+2)C} = \frac{3V}{K+2} \]
Step 4: Check:
Option (A). For \(K = 1\) (no dielectric), \(V' = V\), as expected. Options (B), (C), (D) do not give \(V\) at \(K = 1\).
Final Answer:
Total charge is fixed, capacitance becomes (K + 2)C. \[ \boxed{\text{(A) }\dfrac{3V}{K+2}} \]
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