A) \[2.5\,\,V,\,\,7.5\,\,V\]
B) \[2\,\,V,\,\,8\,\,V\]
C) \[8\,\,V,\,\,2\,\,V\]
D) \[7.5\,\,V,\,\,2.5\,\,V\]
Correct Answer: D
Solution :
Key Idea: The equivalent capacitance for capacitors connected in series is\[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\]. The capacitance \[(C)\] of a parallel plate capacitor is gives by \[{{C}_{A}}=\frac{{{\varepsilon }_{0}}A}{d}\] When dielectric is kept between the plates of \[B\] then \[{{C}_{B}}=\frac{K{{\varepsilon }_{0}}A}{d}=K{{C}_{A}}=3{{C}_{A}}\] The equivalent capacitance for their series combination \[\therefore \] \[C=\frac{{{C}_{A}}\times {{C}_{B}}}{{{C}_{A}}+{{C}_{B}}}\] \[=\frac{{{C}_{A}}\times 3{{C}_{A}}}{{{C}_{A}}+3{{C}_{A}}}=\frac{3}{4}{{C}_{A}}\] \[\therefore \]Net charge\[q=CV=\frac{3}{4}{{C}_{A}}\times 10=7.5{{C}_{A}}\] Hence,\[{{V}_{A}}=\frac{q}{{{C}_{A}}}=7.5\,\,V,\,\,{{V}_{B}}=\frac{q}{{{C}_{B}}}=2.5\,\,V\]
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