IDENTIFY and SET UP:
This problem uses most of the relationships we have discussed for capacitors and dielectrics. (Energy relationships are treated in Example 24.11.) Most of the target variables can be obtained in several ways. The methods used below are a sample; we encourage you to think of others and compare your results.
EXECUTE:
(a) With vacuum between the plates, we use Eq. (24.19) (C=K C_{0}=K \epsilon_{0} \frac{A^{2}}{d x_{0}}=\epsilon \frac{A}{d}) with K = 1:
\begin{aligned}C_{0} &=\epsilon_{0} \frac{A}{d}=\left(8.85 \times 10^{-12} \mathrm{~F} /\mathrm{m}\right) \frac{2.00 \times 10^{-1} \mathrm{~m}^{2}}{1.00 \times 10^{-2}\mathrm{~m}} \\&=1.77 \times 10^{-10} \mathrm{~F}=177 \mathrm{pF}\end{aligned}
(b) From the definition of capacitance, Eq. (24.1) (C=\frac{Q}{V_{a b}}),
\begin{aligned}Q &=C_{0} V_{0}=\left(1.77 \times 10^{-10} \mathrm{~F}\right)\left(3.00\times 10^{3} \mathrm{~V}\right) \\&=5.31 \times 10^{-7} \mathrm{C}=0.531 \mu\mathrm{C}\end{aligned}
(c) When the dielectric is inserted, Q is unchanged but the potential difference decreases to V = 1.00 kV. Hence from Eq. (24.1) (C=\frac{Q}{V_{a b}}), the new capacitance is
\begin{aligned}C &=\frac{Q}{V}=\frac{5.31 \times 10^{-7} \mathrm{C}}{1.00 \times 10^{3}\mathrm{~V}}=5.31 \times 10^{-10} \mathrm{~F} \\&=531 \mathrm{pF}\end{aligned}
(d) From Eq. (24.12) (K=\frac{C}{C_{0}}), the dielectric constant is
\begin{aligned}K &=\frac{C}{C_{0}}=\frac{5.31 \times 10^{-10} \mathrm{~F}}{1.77 \times10^{-10} \mathrm{~F}}=\frac{531 \mathrm{pF}}{177 \mathrm{pF}} \\&=3.00\end{aligned}
Alternatively, from Eq. (24.13) (V=\frac{V_{0}}{K}),
K=\frac{V_{0}}{V}=\frac{3000 \mathrm{~V}}{1000 \mathrm{~V}}=3.00
(e) With K from part (d) in Eq. (24.17) (\epsilon=K \epsilon_{0}), the permittivity is
\begin{aligned}\epsilon &=K \epsilon_{0}=(3.00)\left(8.85 \times 10^{-12} \mathrm{C}^{2} /\mathrm{N} \cdot \mathrm{m}^{2}\right) \\&=2.66 \times 10^{-11} \mathrm{C}^{2} /\mathrm{N}\cdot \mathrm{m}^{2}\end{aligned}
(f) Multiplying both sides of Eq. (24.16) (\sigma_{\mathrm{i}}=\sigma\left(1-\frac{1}{K}\right)) by the plate area A gives the induced charge Q_{\mathrm{i}}=\sigma_{\mathrm{i}} A in terms of the charge Q = σA on each plate:
\begin{aligned}Q_{\mathrm{i}} &=Q\left(1-\frac{1}{K}\right)=\left(5.31 \times 10^{-7}\mathrm{C}\right)\left(1-\frac{1}{3.00}\right) \\&=3.54 \times 10^{-7} \mathrm{C}\end{aligned}
(g) Since the electric field between the plates is uniform, its magnitude is the potential difference divided by the plate separation:
E_{0}=\frac{V_{0}}{d}=\frac{3000 \mathrm{~V}}{1.00 \times 10^{-2} \mathrm{~m}}=3.00 \times 10^{5} \mathrm{~V} / \mathrm{m}
(h) After the dielectric is inserted,
E=\frac{V}{d}=\frac{1000 \mathrm{~V}}{1.00 \times 10^{-2} \mathrm{~m}}=1.00 \times 10^{5} \mathrm{~V} / \mathrm{m}
or, from Eq. (24.18) (E=\frac{\sigma}{\epsilon}),
\begin{aligned}E &=\frac{\sigma-\sigma_{\mathrm{i}}}{\epsilon_{0}}=\frac{Q-Q_{\mathrm{i}}}{\epsilon_{0} A} \\&=\frac{(5.31-3.54) \times 10^{-7} \mathrm{C}}{\left(8.85 \times10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}\right)\left(2.00 \times10^{-1} \mathrm{~m}^{2}\right)} \\&=1.00 \times 10^{5} \mathrm{~V} / \mathrm{m}\end{aligned}
or, from Eq. (24.14) (E_{0}=\frac{\sigma}{\epsilon_{0}} \quad E=\frac{\sigma-\sigma_{\mathrm{i}}}{\epsilon_{0}}),
E=\frac{E_{0}}{K}=\frac{3.00 \times 10^{5} \mathrm{~V} / \mathrm{m}}{3.00}=1.00 \times 10^{5} \mathrm{~V} / \mathrm{m}
EVALUATE: Inserting the dielectric increased the capacitance by a factor of K = 3.00 and reduced the electric field between the plates by a factor of 1/K = 1/3.00. It did so by developing induced charges on the faces of the dielectric of magnitude Q(1-1 / K)=Q(1-1 / 3.00)=0.667 Q.
