TS EAMCET · Physics · Capacitance
As shown in the figure, a dielectric of constant K is placed between the plates of a parallel plate capacitor and is charged to a potential V using a battery. If the dielectric is pulled out after disconnecting the battery from the capacitor, the final potential difference across the plates of the capacitor is
- A \(\left(1+\frac{1}{\mathrm{~K}}\right) 2 \mathrm{~V}\)
- B 2 KV
- C \(\frac{2 \mathrm{~V}}{\left(1+\frac{1}{\mathrm{~K}}\right)}\)
- D \(\frac{\mathrm{V}}{2}\left(1+\frac{1}{\mathrm{~K}}\right)\)
Answer & Solution
Correct Answer
(C) \(\frac{2 \mathrm{~V}}{\left(1+\frac{1}{\mathrm{~K}}\right)}\)
Step-by-step Solution
Detailed explanation
\(C_i = \frac{2KC_0}{K+1}\) \(Q = C_i V = \frac{2KC_0 V}{K+1}\) \(C_f = C_0\) \(V_f = \frac{Q}{C_f} = \frac{2KV}{K+1}\) \(V_f = \frac{2V}{1+\frac{1}{K}}\)
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