question_answer

Prove that during the charging of a parallel plate capacitor, the rate of change of charge on each plate equals \[{{\varepsilon }_{0}}\] times the rate of change of electric flux  \[{{\phi }_{E}}\] linked with it, Also, write the  name given to the term \[{{\varepsilon }_{0}}\frac{d{{\phi }_{E}}}{dt}.\]

Answer:

Charge on each plate of a parallel plate capacitor can be expressed as \[q(t)=\sigma (t)A\] Also,     \[\sigma (t)={{\varepsilon }_{0}}E(t)\]     \[\Rightarrow q(t)={{\varepsilon }_{0}}AE(t)\] where, \[\sigma (t)\]=  instantaneous charge per unit area E (t) = electric field strength As,        E(t) A = electric flux \[{{\phi }_{E}}(t)\] \[\therefore \]      \[q(t)={{\varepsilon }_{0}}{{\phi }_{E}}(t)\] Therefore, rate of change of charge \[\frac{dq(t)}{dt}={{\varepsilon }_{0}}\frac{d{{\phi }_{E}}(t)}{dt}\] \[\therefore \] Rate of change of charge \[={{\varepsilon }_{0}}\times \]rate of change of electric flux\[({{\phi }_{E}})\]. The quantity \[{{\varepsilon }_{0}}\frac{d{{\phi }_{E}}(t)}{dt}\] is named as displacement current.