Saved Bookmarks
| 1. |
A parallel plate capacitor with air between the plates has a capacitance of 9 pF. The separation between its plates is ‘d'. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant K_1 =3and thickness d/3 while the other one has dielectric constant K_2 = 6 and thickness2/3d. Capacitance of the capacitor is now |
|
Answer» `1.8 pF` When capacitor is filled with two DIELECTRICS, the NEW capacitance `C. = (epsi_0A)/((d_1/K_1 + d_1/k_2))=(epsi_0A)/(((d//3)/3+ (2D//3)/6))=(epsi_0A)/(((2d)/9)) = 9/2 (epsi_0A)/d = 9/2 xx 9 pF= 40.5 pF` |
|