Question

Two parallel-plate air capacitors, each of capacitance $C$, were connected in series to a battery with emf $ϵ$. Then one of the capacitors was filled up with uniform dielectric with permittivity $ϵ$. How many times did the electric field strength in that capacitor decrease? What amount of charge flows through the battery?

Answer 1

From the symmetry of the problem, the voltage across each capacitor, $△\phi =\xi /2$ and charge on each capacitor $q=C\xi /2$ in the absence of dielectric.

Now when the dielectric is filled up in one of the capacitors, the equivalent capacitance of the system,

$C_0^{\prime }=\frac{Cϵ}{1+ϵ}$

and the potential difference across the capacitor, which is filled with dielectric,

$△{\phi }^{\prime }=\frac{q^{\prime }}{ϵC}=\frac{Cϵ}{(1+ϵ)}\frac{\xi }{Cϵ}=\frac{\xi }{(1+ϵ)}$

But $\phi \propto E$

So, as $\phi$ decreases $\frac{1}{2}(1+ϵ)$ times, the field strength also decreases by the same factor and flow of charge, $△q=q^{\prime }-q$

$=\frac{Cϵ}{(1+ϵ)}\xi -\frac{C}{2}\xi =\frac{1}{2}C\xi \frac{(ϵ-1)}{(ϵ+1)}$.