MHT CET · Physics · Capacitance
Two identical parallel plate air capacitors are connected in series to a battery of e..m.f. \({ }^{\prime} \mathrm{V}^{\prime}\). If one of the capacitor is inserted in liquid of dielectric constant '\(\mathrm{K}^{\prime}\), then potential difference of the other capacitor will become
- A \(\frac{\mathrm{K}}{\mathrm{V}(\mathrm{K}+1)}\)
- B \(\frac{\mathrm{KV}}{\mathrm{K}+1}\)
- C \(\frac{\mathrm{K}+1}{\mathrm{KV}}\)
- D \(\frac{\mathrm{K}}{\mathrm{V}(1-\mathrm{K})}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{KV}}{\mathrm{K}+1}\)
Step-by-step Solution
Detailed explanation
Let the capacitance of capacitor \(C_{2}\) be \(C\). Thus capacitance of capacitor \(C_{1}\) is KC.
Potential across the capacitor \(\mathrm{C}_{2}, \mathrm{~V}_{2}=\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}} \mathrm{~V}\)
\(\therefore \mathrm{V}_{2}=\frac{\mathrm{KC}}{\mathrm{C}_{1}+\mathrm{KC}} \mathrm{V}\)
\(\Longrightarrow \mathrm{V}_{2}=\frac{\mathrm{KV}}{\mathrm{K}+1}\)
Potential across the capacitor \(\mathrm{C}_{2}, \mathrm{~V}_{2}=\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}+\mathrm{C}_{2}} \mathrm{~V}\)
\(\therefore \mathrm{V}_{2}=\frac{\mathrm{KC}}{\mathrm{C}_{1}+\mathrm{KC}} \mathrm{V}\)
\(\Longrightarrow \mathrm{V}_{2}=\frac{\mathrm{KV}}{\mathrm{K}+1}\)
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