Question

When dielectric of dielectric k is introduced between the plates of a parallel plate capacitor and battery is connected across the capacitor electric feild decreases k times. Explain why

Answer 1

The electric field when there is medium between the charges:
$$\begin{gathered} E=\frac{1}{4\pi \in }\frac{Q^2}{r^2}...\left(1\right) \\ \text{where},\in =\text{electrical permittivity of medium} \\ r=\text{distance between the charge} \\ Q=\text{charge} \\ \text{As, dielectric contant is given as:}K=\frac{\in }{{\in }_0} \\ \Rightarrow \in =K{\in }_0 \\ \text{Putting this value in eq}.\left(1\right) \\ E=\frac{1}{4\pi K{\in }_0}\frac{Q^2}{r^2}=\frac{1}{K}\frac{1}{4\pi {\in }_0}\frac{Q^2}{r^2}=\frac{E_0}{K} \end{gathered}$$
Therefore, the electric field between the plates decreases K times.