Saved Bookmarks
| 1. |
Deduce the expression for the electrostatic energy stored in a capacitor of capacitance 'C' and having charge 'Q'. How will the (i) energy stored and (ii) the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant 'K' ? |
|
Answer» SOLUTION :Potential difference between the plates of capacitor V = q/C Work done to add additional charge dg on the capacitor ` dw = V xx dq` ` therefore ` Total energy stored in the capacitor ` U = INT d w = int_0^Q dq = 1/2 (Q^2)/(C )` When battery is disconnected, (i)Energy stored will be DECREASED or energy stored = 1/K TIMES the initial energy (ii) Electric field WOULD decrease or E. = E/K |
|