## University Physics with Modern Physics (14th Edition)

(a) $Q_o = 0.36 \mathrm{~n C}$ (b) $Q_K = 0.97 \mathrm{~n C}$
(a) the capacitance between the two plates occur due to the accumulation of the charges on the plates because of the potential and for the air, it is given by $$Q_o = C_oV$$ Where $V_{ab}$ is the potential between the two plates and is calculated by $$V = Ed = ( 3 \times 10^{4} \,text{V/m})(0.0015 \,\text{m}) = 45 \,\text{V}$$ Let us substitute the values $C$ and $V$ to get the charge on each plate $$Q_o = C_oV = (8\times 10^{-12} \,\text{F})(45 \,\text{V}) = \boxed{0.36 \mathrm{~n C}}$$ (b) After the dielectric material with a dielectric constant is added $K = 2.70$ the charges will be given by $$Q_K = K C_oV = 2.70 (8\times 10^{-12} \,\text{F})(45 \,\text{V}) = \boxed{0.97 \mathrm{~n C}}$$