Saved Bookmarks
| 1. |
Figure 25-22 shows a parallel plate capacitor of plate area A and plate separation d. A potential difference V_0 is applied between the plates by connecting a battery between them. The battery is then disconnected and a dielectric slab of thickness b and dielectric constant k is placed between the plates as shown. Assume A=115 cm^2,d=1.2cm, V_0=85.5 V, b=0.780 cm and k=2.61 What is the capacitance C_0 before the dielectric slab is inserted? (b) What free charge appears on the plates? (c ) What is the electrci field E_0 in the gaps between the plates and teh dielectric slab? (d) What is the electric field E_1 in the dielectric slab? |
|
Answer» Solution :a) From Eq.25-9 we have `C_0=(epsilon_0A)/d=((8.85 times 10^-12 F//m)(115 times 10^-4 m^2))/(1.24 times 10^-2 m)` `=8.21 times 10^-12 F =8.21 pF` b) From Eq.25-1 `q=C_0V_0=(8.21 times 10^-12 F)(85.5V)` `=7.02 times 10^-10 C=702 pC` Because the BATTERY was DISCONNECTED before the slab was INSERTED. the free charge is unchanged. |
|