• # 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 ${{\varepsilon }_{0}}\frac{d{{\phi }_{E}}}{dt}.$

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.