Saved Bookmarks
| 1. |
A slab of material of dielectric constant K hasthe same area as the plates of a parallel capacitor, but hasa thickness ((3)/(4) d), where d is the separation of the plates. How is the capacitancechangedwhen the slab is inserted between the plates |
|
Answer» Solution : ` V = E_o ((d)/(4))+(E_o)/(K)((3d)/(4)) = E_o d ((K+3)/(4K)) ` ` V = V_o (( K + 3 )/(4K)) ` ` C = (Q_o)/(V)= (4K)/(K + 3) (Q_o)/(V_o) = (4 K )/(K + 3 ) CO` |
|