A) \[\frac{3V}{K+2}\]
B) \[\frac{3V}{K}\]
C) \[\frac{V}{K+2}\]
D) \[\frac{V}{K}\]
Correct Answer: A
Solution :
\[{{q}_{1}}=2CV,\] \[{{q}_{2}}=CV\] Now condenser of capacity C is filled with dielectric K, therefore C2 = KC As charge is conserved \ \[{{q}_{1}}+{{q}_{2}}=(C_{2}^{'}+2C)V'\] Þ \[V'=\frac{3CV}{(K+2)C}=\frac{3V}{K+2}\]
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