Solution:
The relation between the potential difference and the electric field between the plates of the parallel plate capacitor is given by :
[tex]$E=\frac{V}{D}$[/tex]
Differentiating on both the sides with respect to time, we get
[tex]$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$[/tex]
Therefore, the rate of the electric field changes between the plates of the parallel plate capacitor is given by :
[tex]$\frac{dE}{dt}=\frac{1}{D}\frac{dV}{dt}$[/tex]
[tex]$=\frac{1}{1.4 \times 10^{-3}} \times 110$[/tex]
[tex]$=7.85 \times 10^4$[/tex] V/m-s