A) The charge on the condenser will be doubled
B) The charge on the condenser will be reduced to half
C) The P.D. across the condenser will be \[320\ volts\]
D) The P.D. across the condenser will be \[45\ volts\]
Correct Answer: D
Solution :
If nothing is said, it is considered that battery is disconnected. Hence charge remain the same Also \[{{V}_{air}}=\frac{\sigma }{{{\varepsilon }_{0}}}\times d\] and \[{{V}_{medium}}=\frac{\sigma }{{{\varepsilon }_{0}}}(d-t+\frac{t}{k})\] Þ\[\frac{{{V}_{m}}}{{{V}_{a}}}=\frac{(d-t+\frac{t}{k})}{d}\] Þ \[\frac{{{V}_{m}}}{120}=\frac{(8-6+\frac{6}{6})}{8}\] Þ Vm = 45V
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