Question

if we insert dielectric in capacitor keeping the capacitor connected to battery then the elevtric field between the capacitor is decreased by the factor k the how the potential i.i E.d remain constan

Answer 1

The potential difference remains constant because the battery maintains a fixed voltage. Since the distance between the plates is also constant, the electric field must remain constant ($E = V/d$). The premise that the net electric field decreases by a factor $k$ is incorrect when the capacitor is connected to a battery.

Answer 2

When a capacitor is connected to a battery, the potential difference ($V$) across its plates is held constant by the battery. The electric field ($E$) between the parallel plates is related to the potential difference and the distance between the plates ($d$) by the formula $V = E \times d$. Since $V$ is constant and $d$ is constant, the net electric field $E$ must also remain constant. The dielectric material polarizes, creating an induced electric field ($E_{ind}$) that opposes the field due to free charges ($E_{free}$). The net field is $E = E_{free} - E_{ind}$. The dielectric constant $k$ is defined such that $E = E_{free} / k$. Because the net field $E$ is constant, the field due to free charges ($E_{free}$) must increase by a factor of $k$ ($E_{free} = k \times E$) to compensate for the induced field. Therefore, the potential difference remains constant, and the net electric field also remains constant, contrary to the premise that it decreases by a factor $k$.