Saved Bookmarks
| 1. |
A parallel plate capacitor has plates of area 4 m^(2) separated by distance of 0.5 mm. The capacitor is connected across a cell of emf 100 V. (a) Find the capacitance, charge and energy stored in the capacitor. (b) A dielectric slab of thickness 0.5 mm is inserted inside this capacitor after it has been disconnected from the cell. Find the answers to part(a) if K = 3. |
|
Answer» Solution :(a) ` C_0 = (in _0 A)/( d) =( 8.85xx 10 ^(-22) xx4)/( 0.5xx 10 ^(-5)) = 7.08xx 10 ^(-2)muF` ` Q_0 =C_0 V_0 =( 7.08xx 10 ^(-2) xx 100 ) muC= 7.08 mu C ` ` U_0 = (1)/(2) C_0 V_0 ^(2) = 3.54 xx10 ^(-4) J` (b) As the cell has been disconnected ` Q= Q_0` ` C= (Kin _0A)/( d) =KC_0 = 0.2124mu F ` `V= (Q)/(C) =(Q_0)/(KC_0) =( V_0)/( K) =(100)/( 3)V ` ` U = (Q_0^(2))( 2KC_0)= (U_0)/(K) =118 xx 10 ^(-6) J` |
|