A) 2.5V, 7.5 V
B) 2V, 8V
C) 8V, 2V
D) 7.5V, 2.5V
E) 3V, 2V
Correct Answer: D
Solution :
\[{{C}_{A}}=\frac{{{\varepsilon }_{0}}A}{d},{{C}_{B}}=\frac{3{{\varepsilon }_{0}}A}{d}=3{{C}_{A}}\] \[C=\frac{{{C}_{A}}\times {{C}_{B}}}{{{C}_{A}}+{{C}_{B}}}\] \[=\frac{{{C}_{A}}\times 3{{C}_{A}}}{{{C}_{A}}+3{{C}_{A}}}\] \[=\frac{3}{4}{{C}_{A}}\] \[\therefore \]Net charge \[q=CV=\frac{3}{4}{{C}_{A}}\times 10=7.5\,{{C}_{A}}\] Hence, \[{{V}_{A}}=\frac{q}{{{C}_{A}}}=7.5V\] and \[{{V}_{B}}=\frac{q}{{{C}_{B}}}=2.5V\]
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