Answer:
[tex]I=V\epsilon_o (K-1) * \frac{r^2}{d*\triangle t}[/tex]
Explanation:
Since a slab is inserted between the plates, so we consider them as two capacitors attached in Parallel. One with dielectric and one without dielectric.
Equation will become:
Total Capacitance=Capacitance With dielectric+ Capacitance without Dielectric
[tex]C=\epsilon_oK*a*\frac{r}{d}+ \epsilon_o(r-a)\frac{r}{d}[/tex]
Where:
a is the distance at which slab is added between the plates.
Rearranging the above equation:
[tex]C=\epsilon_o*\frac{r}{d} [Ka+ (r-a)][/tex]
Charge on the capacitor Q is given by:
Q=CV
Current "I" is given by:
[tex]I=\frac{dQ}{dt}[/tex]
Now,
[tex]I=\frac{d(CV)}{dt} \\\\I=\frac{Vd(C)}{dt} \\I=\frac{Vd[\epsilon_o*\frac{r}{d} [Ka+ (r-a)]]}{dt} \\[/tex]
Taking derivative:
[tex]I=V\epsilon_o}*\frac{r}d}[\frac{d(Ka)}{dt}+\frac{dr}{dt}-\frac{d(a)}{dt}]\\I=V\epsilon_o * \frac{r}{d} (K-1)\frac{d(a)}{dt}[/tex] dr/dt=0 (r is constant)
In the above equation, d(a)/dt is the speed which is constant.
Speed= Distance/time
d(a)/dt= r/Δt
Final Equation will become:
[tex]I=V\epsilon_o * \frac{r}{d} (K-1)\frac{r}{\triangle t} \\I=V\epsilon_o (K-1) * \frac{r^2}{d*\triangle t}[/tex]
