InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The two plates of a parallel plate capacitor are 4 mm apart . A slab of dielectric constant 3 and thickness 3 mm is introduced between the plates with its face parallel to them . The distance between the plates is so adjusted that the capacity of the capacitor becomes 2/3 rd of its original value . What is the new distance the plates ? |
|
Answer» Solution :Here d = 4 mm , t = 3 mm , K = 3 and `C. = (2)/(3)` C . Let new distance between the capacitor PLATES be d. mm . Then `C = (in_0 A)/(d) = (in_(0) A)/(4 mm) … (i) and C.= (2)/(3) C (in_(0) A)/(d. - t (1 - (t)/(K))) = (in_(0) A)/(d.- 3 (1 - (1)/(3))) = (in_(0) A)/((d.-2)) .... (ii)` Dividing (i)by (ii) , we get `(3)/(2) = (d.-2)/(4 mm) implies d. = 8` mm |
|