question_answer

The capacitors of capacitance 4\[\mu F\] and 6\[\mu F\] are connected in series. A potential difference of 500V is applied to the outer plates of two capacitor system. Then the charge on each plate of each capacitor is numerically:

A)  6000 \[\mu C\]                

B)  1200\[\mu C\]

C)  12000\[\mu C\]               

D)  6000\[\mu C\]

Correct Answer: A

Solution :

                 The circuit diagram of the set up is as shown. Equivalent capacitance in series is \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\] \[=\frac{1}{4}+\frac{1}{6}=\frac{5}{12}\] \[\Rightarrow \]   \[C=\frac{12}{5}\mu F\] Also, charge = capacitance \[\times \] voltage \[Q=\left( \frac{12}{5}\times 500 \right)\mu C\] \[Q=1200\,\mu C\]