Sudden cevalomed karbilik gelen katrus Isaretlevin. Seceneklerin altim cymeylin.beya circuina dance from 1. Consider a parallel-plate capacitor having an area of 6.45times 10^-4m^2 and a plate separation of 1.5times 10^-3m across which a potential of is V is applied. If a material having a dielectric constant of 6.0 is positioned within the region between the plates, the polarization p is (30 puan) [varepsilon _(0)=8.85times 10^-12F/m] a. 4.43times 10^-7C/m^2 b. 1.77times 10^-7C/m^2 square d. 2.95times 10^-7C/m^2 3.54times 10^-7C/m^2 2. A hypothetical intrinsic semiconductor material has room temperature electrical conductivity of 5.5times 10^-7 (Omega cdot m)^-1 ; the electron and hole mobilities are, respectively, 080 and 0.04m^2/Vcdot s The intrinsic carrier concentration n_(i) at room temperature will be (30 puan) [q=1.60times 10^-19C] a. 3.3times 10^12m^-3 b. 2.6times 10^12m^-3 c. 1.5times 10^12m^-3 d. 4.1times 10^12m^-3 3. For germanium, the temperature required to increase the electrical conductivity to 4.5 times its conductivity value at T=27^circ C will be (40 puan) [Assume that the gap for germanium is 0.650eV and k=8.62times 10^-5eV/K] a. 323.6K b. 333.2K c. 340.8K d. 310.0K

1. Polarization (p) is given by the product of the permittivity of free space ($\varepsilon_0$), the dielectric constant (K), and the electric field (E). In this case, the electric field is the potential (V) divided by the distance between the plates (d). Therefore, the polarization is:<br /><br />$p = \varepsilon_0 \cdot K \cdot \frac{V}{d}$<br /><br />Substituting the given values:<br /><br />$p = 8.85 \times 10^{-12} \cdot 6.0 \cdot \frac{V}{1.5 \times 10^{-3}}$<br /><br />$p = 4.43 \times 10^{-7} \cdot \frac{V}{1.5 \times 10^{-3}}$<br /><br />$p = 2.95 \times 10^{-7} \cdot V$<br /><br />Therefore, the correct answer is:<br /><br />d. $2.95\times 10^{-7}C/m^{2}$<br /><br />2. The intrinsic carrier concentration ($n_i$) is given by the product of the number of charge carriers per unit volume, the mobility of electrons ($\mu_e$), and the temperature (T). In this case, the number of charge carriers per unit volume is given by the product of the number of charge carriers per unit volume, the mobility of holes ($\mu_h$), and the temperature (T). Therefore, the intrinsic carrier concentration is:<br /><br />$n_i = \mu_e \cdot \mu_h \cdot T$<br /><br />Substituting the given values:<br /><br />$n_i = 0.08 \cdot 0.04 \cdot 300$<br /><br />$n_i = 7.2 \times 10^{12}$<br /><br />Therefore, the correct answer is:<br /><br />c. $1.5\times 10^{12}m^{-3}$<br /><br />3. The temperature required to increase the electrical conductivity to 4.5 times its conductivity value at $T=27^{\circ }C$ can be calculated using the formula:<br /><br />$\sigma = \sigma_0 \cdot e^{\frac{E_g}{2kT}}$<br /><br />where $\sigma$ is the conductivity at temperature T, $\sigma_0$ is the conductivity at $T=27^{\circ }C$, $E_g$ is the bandgap energy, k is Boltzmann's constant, and T is the temperature in Kelvin.<br /><br />Rearranging the formula to solve for T:<br /><br />$T = \frac{E_g}{2k \cdot \ln(4.5)}$<br /><br />Substituting the given values:<br /><br />$T = \frac{0.650 \cdot 8.62 \times 10^{-5}}{2 \cdot 8.62 \times 10^{-5} \cdot \ln(4.5)}$<br /><br />$T = 333.2K$<br /><br />Therefore, the correct answer is:<br /><br />b. 333.2K