(a) Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the…

In a parallel combination the potential difference is the same across the capacitors. On insertion of the dielectric slab the capacitance of each capacitor becomes K (dielectric constant) times, i.e. $\displaystyle \mathrm{C}^{\prime}=K \mathrm{C}$. Hence the final charge on each capacitor becomes K times: $\displaystyle q^{\prime}=\mathrm{C}^{\prime} V=K \mathrm{C} V=K q$, so $\displaystyle q_{1}^{\prime}=K \mathrm{C}_{1} V=K q_{1}$ and $\displaystyle q_{2}^{\prime}=K \mathrm{C}_{2} V=K q_{2}$.

Energy stored by a capacitor $\displaystyle U=\dfrac{1}{2} \mathrm{C} V^{2}$. As the capacitance becomes K times, the energy stored by each capacitor also increases by a factor K, i.e. $\displaystyle U^{\prime}=K U$, so $\displaystyle U_{1}^{\prime}=\dfrac{1}{2} K \mathrm{C}_{1} V^{2}=K U_{1}$ and $\displaystyle U_{2}^{\prime}=\dfrac{1}{2} K \mathrm{C}_{2} V^{2}=K U_{2}$.

$\displaystyle \vec{v}_{d}=\dfrac{-e \vec{E} \tau}{m}$, where $\displaystyle \tau$ is the average time between two successive collisions, which is constant. The distance covered by electrons between two successive collisions is almost the same, hence $\displaystyle S=\dfrac{1}{2} a t^{2}$ where $\displaystyle a$ = constant and $\displaystyle s$ = almost constant, so $\displaystyle t$ is constant.

Alternatively: At any instant some of the electrons would have spent time more than $\displaystyle \tau$ and some less than $\displaystyle \tau$. Thus averaging over all the times we get a constant time. This way we get an average velocity independent of time.