### Related Formula By Gauss's Law, the total flux emitted by a point charge q$q$ through a completely enclosing symmetric cube container surface is: Phitexttotal = fracqepsilon0 $\Phi{\text{total}} = \frac{q}{\epsilon{0}} $ ### Core Logic Assuming the charge resides at a symmetric center distance fraca2$\frac{a}{2}$ relative to the loop face [cite: 784, 786], this square loop represents one of the six identical faces of an enclosing cube system. Thus, the flux passing through the entire square loop face is[cite: 775, 776]: Phitextsquare = frac16 Phitexttotal = fracq6epsilon0$\Phi{\text{square}} = \frac{1}{6} \Phi{\text{total}} = \frac{q}{6\epsilon{0}}$ ### Step 1: Symmetric Partitioning As shown in the solution schematic

, the square face can be divided into 8 identical symmetric right-angled triangle sections by drawing its diagonals and medians[cite: 771, 777]. Each individual part intercepts an equal portion of the flux field : Phitextpart = frac18 Phitextsquare = frac18 left(fracq6epsilon_0 ight) = fracq48epsilon_0$\Phi{\text{part}} = \frac{1}{8} \Phi{\text{square}} = \frac{1}{8} \left(\frac{q}{6\epsilon_{0}} ight) = \frac{q}{48\epsilon_{0}}$ The shaded region covers exactly 5 of these individual triangle parts [cite: 779, 780]: Phi_textshaded = 5 times Phi_textpart = frac548 times fracqepsilon_0$\Phi_{\text{shaded}} = 5 \times \Phi_{\text{part}} = \frac{5}{48} \times \frac{q}{\epsilon_{0}}$ Comparing this result with the given expression frac5p times frac1epsilon_0$\frac{5}{p} \times \frac{1}{\epsilon_{0}}$ , we find: p = 48$p = 48$ ### Pattern Recognition Exploit geometric symmetry to break solid angles down into equal fractions, avoiding complex surface integration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula The electric field E$E$ between the plates of a parallel plate capacitor is given by : E = fracVd$E = \frac{V}{d}$ The electrostatic energy density U$U$ in a medium is given by: U = frac12 epsilon_0 E^2$U = \frac{1}{2} \epsilon_0 E^2$ ### Core Logic Given values [cite: 662, 664, 665]: * Capacitance, C = 1 \ mutextF$C = 1 \ \mu\text{F}$ * Potential difference, V = 20 text V$V = 20 \text{ V}$ * Plate separation, d = 1 \ mutextm = 10^-6 text m$d = 1 \ \mu\text{m} = 10^{-6} \text{ m}$ First, calculate the electric field E$E$ : E = frac2010^-6 = 20 times 10^6 text V/m$E = \frac{20}{10^{-6}} = 20 \times 10^6 \text{ V/m}$ Now, calculate the energy density U$U$ using epsilon_0 approx 8.85 times 10^-12 text F/m$\epsilon_0 \approx 8.85 \times 10^{-12} \text{ F/m}$ : U = frac12 times (8.85 times 10^-12) times (20 times 10^6)^2$U = \frac{1}{2} \times (8.85 \times 10^{-12}) \times (20 \times 10^6)^2$ U = frac12 times 8.85 times 10^-12 times 400 times 10^12$U = \frac{1}{2} \times 8.85 \times 10^{-12} \times 400 \times 10^{12}$ U = 8.85 times 200 = 1770 text J/m^3 = 1.77 times 10^3 text J/m^3$U = 8.85 \times 200 = 1770 \text{ J/m}^3 = 1.77 \times 10^3 \text{ J/m}^3$ Rounding to the nearest matching option gives 1.8 times 10^3 text J/m^3$1.8 \times 10^3 \text{ J/m}^3$. ### Pattern Recognition Notice that energy density is completely independent of the total capacitance value C$C$ if the voltage and spacing are directly provided. Always check for redundant parameters included to distract candidates. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula The potential energy of an electric dipole aligned at an \angle theta$\theta$ inside a uniform electric field is given by: U = -p E cos theta$U = -p E \cos \theta$ The work done by an external agent to rotate the dipole equals its change in potential energy: W = Delta U = -pE (cos theta_f - cos theta_i)$W = \Delta U = -pE (\cos \theta_f - \cos \theta_i)$ ### Core Logic Given parameters [cite: 805, 806]: * Dipole moment, p = 6 times 10^-6 text Cm$p = 6 \times 10^{-6} \text{ Cm}$ * Electric field, E = 10^6 text V/m$E = 10^6 \text{ V/m}$ * Initial \angle (\parallel state), theta_i = 0^circ$\theta_i = 0^{\circ}$ * Final \angle (opposite state), theta_f = 180^circ$\theta_f = 180^{\circ}$ Substitute the values into the work equation [cite: 807, 808]: W = -pE (cos 180^circ - cos 0^circ)$W = -pE (\cos 180^{\circ} - \cos 0^{\circ})$ W = -pE (-1 - 1) = 2 p E quad text$W = -pE (-1 - 1) = 2 p E \quad \text{}$ W = 2 times (6 times 10^-6) times 10^6 = 12 text J quad text$W = 2 \times (6 \times 10^{-6}) \times 10^6 = 12 \text{ J} \quad \text{}$ ### Pattern Recognition Rotating a dipole from its most stable configuration (\parallel, theta=0^circ$\theta=0^{\circ}$) to its most unstable position (anti-\parallel, theta=180^circ$\theta=180^{\circ}$) always requires a maximum work value of exactly 2pE$2pE$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Core Logic Mapping electrostatics equations via Gauss\'s law applications : * (A) Inside a shell, enclosed charge is zero implies E = 0$\implies E = 0$ (III) . * (B) Near an infinite sheet, E = fracsigma2epsilon_0$E = \frac{\sigma}{2\epsilon_0}$ (II) . * (C) Outside a shell, E = frackQr^2 = fracsigma R^2epsilon_0 r^2$E = \frac{kQ}{r^2} = \frac{\sigma R^2}{\epsilon_0 r^2}$ (IV) . * (D) Between opposite sheets, fields add up: fracsigma2epsilon_0 + fracsigma2epsilon_0 = fracsigmaepsilon_0$\frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$ (I) . Hence, the proper combination sequence is (A)-(III), (B)-(II), (C)-(IV), (D)-(I). ### Chapter Mix Class 12 Physics: Electrostatics