MHT CET · Physics · Capacitance
Two identical parallel plate air capacitors are connected in series to a battery of emf ' \(\mathrm{V}\) '. If one of the capacitor is inserted in liquid of dielectric constant ' \(\mathrm{K}\) ' then, potential difference of the other capacitor will become
- A \(\frac{\mathrm{K}-1}{\mathrm{KV}}\)
- B \(\frac{\mathrm{K}+1}{\mathrm{KV}}\)
- C \(\left(\frac{\mathrm{KV}}{\mathrm{K}+1}\right)\)
- D \(\frac{\mathrm{KV}}{\mathrm{K}-1}\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{\mathrm{KV}}{\mathrm{K}+1}\right)\)
Step-by-step Solution
Detailed explanation
Let the capacitance of capacitors are C initially.
So, afterward \(\mathrm{C}_1=\mathrm{KC}\) and \(\mathrm{C}_2=\mathrm{C}\)
p.d. across capacitor \(\mathrm{C}_2\) is
\(\mathrm{V}_2=\frac{\mathrm{C}_1 \mathrm{~V}}{\mathrm{C}_1+\mathrm{C}_2}=\frac{\mathrm{KCV}}{\mathrm{C}+\mathrm{KC}}=\frac{\mathrm{KV}}{\mathrm{K}+1}\)
So, afterward \(\mathrm{C}_1=\mathrm{KC}\) and \(\mathrm{C}_2=\mathrm{C}\)
p.d. across capacitor \(\mathrm{C}_2\) is
\(\mathrm{V}_2=\frac{\mathrm{C}_1 \mathrm{~V}}{\mathrm{C}_1+\mathrm{C}_2}=\frac{\mathrm{KCV}}{\mathrm{C}+\mathrm{KC}}=\frac{\mathrm{KV}}{\mathrm{K}+1}\)
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