A) \[2.5\,V,7.5\,V\]
B) \[2\,V,8V\]
C) \[8\,V,2V\]
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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