Answer
a) decreases
b) decreases
c) increases
d) decreases
Work Step by Step
(a) We know that $ E=\frac{E_{\circ}}{K}$. This equation shows that the electric field is decreased by a factor $ K $, in the presence of a dielectric constant $ K $.
(b) We know that $ V=\frac{V_{\circ}}{K}$. Here $ V $ is the potential difference in the presence of a dielectric constant $ K $ and $ V_{\circ}$ is the electric potential in the absence of the dielectric. Thus, the potential difference between the plates of the capacitor decreases by a factor $ K $.
(c) The capacitance of a capacitor increases by a factor $ K $ when a dielectric is present between the two plates because $ C=KC_{\circ}$.
(d) We know that $ U=\frac{1}{2}QV $
$ U=\frac{1}{2}CV^2$
$\implies U=\frac{1}{2}Q\frac{V_{\circ}}{K}$
$\implies U=\frac{U_{\circ}}{K}$
Thus, the energy stored in a capacitor decreases by a factor $ K $.
