question_answer

Parallel plate capacitors \[{{C}_{1}}\] & \[{{C}_{2}}\] have the same area with \[{{C}_{1}}\]having half the plate separation as \[{{C}_{2}}\] . \[{{C}_{1}}\] is inserted wholly inside \[{{C}_{2}}\] as shown with their plates joined as shown. The capacitance across terminals AB will be

A)  \[{{C}_{1}}+{{C}_{2}}\]                   

B)  \[\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}\]

C)  \[{{C}_{1}}\]                                   

D)  \[{{C}_{2}}\]

Correct Answer: C

Solution :

 Since the plates are joined. Potential of A is equal to potential of C and potential of B is equal to potential of D. Hence potential difference between AB is equal to  potential difference between CD. \[\therefore \] Capacitance across AB is equal to capacitance across CD which is \[{{C}_{1}}\]