Question

A thin metal plate P is inserted between the plates of a parallel-plate capacitor of capacitance C in such a way that its edges touch the two plates (figure 31-Q2). The capacitance now becomes
(a) C/2
(b) 2 C
(c) 0
(d) [infinity].
Figure

Answer 1

Answer:

d........... ...,. ....,....

Answer 2

(d) The capacitance now becomes infinity.

Explanation:

• It is to be noted that the Capacitance C=Q/V1−V2 (V1−V2) is the potential variance between the plates of the capacitor’s plates. Also, when plates are connected by a wire than V1=V2 so V1−V2=0  

C= Q/V1−V2 = Q/0=∞ [infinity].

• Capacitance is the system’s energy for storing electric charge. It is the ratio of the electric charge’s change in a system to the corresponding change. This is in its electric potential.