### Related Formula sigma propto frac1R_textcurv$\sigma \propto \frac{1}{R_{\text{curv}}}$ where, sigma$\sigma$ = surface charge density R_textcurv$R_{\text{curv}}$ = local radius of curvature at that point on the conductor's surface ### Core Logic On an irregular-shaped charged metallic conductor in electrostatic equilibrium: - The electric potential is identical at all points on the surface. - However, the surface charge density sigma$\sigma$ is not uniform. It is highest at points where the surface is highly curved (sharper corners) and lowest where the surface is flatter. Analyzing the radii of curvature (R_textcurv)$(R_{\text{curv}})$ from the figure: - Point 1 is the sharpest corner (smallest radius of curvature): (R_textcurv)_1$(R_{\text{curv}})_1$ - Point 3 is less sharp: (R_textcurv)_3$(R_{\text{curv}})_3$ - Points 2 and 4 are symmetric flat regions of equal curvature: (R_textcurv)_2 = (R_textcurv)_4$(R_{\text{curv}})_2 = (R_{\text{curv}})_4$ Therefore, we have: (R_textcurv)_1 < (R_textcurv)_3 < (R_textcurv)_2 = (R_textcurv)_4$(R_{\text{curv}})_1 < (R_{\text{curv}})_3 < (R_{\text{curv}})_2 = (R_{\text{curv}})_4$ Using the inverse relationship sigma propto frac1R_textcurv$\sigma \propto \frac{1}{R_{\text{curv}}}$: sigma_1 > sigma_3 > sigma_2 = sigma_4$\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$ ### Step 1: Verification of Statements - Statement (A) sigma_1 > sigma_3$\sigma_1 > \sigma_3$; sigma_2 = sigma_4$\sigma_2 = \sigma_4$ is **Correct**. - Statement (B) sigma_1 > sigma_2$\sigma_1 > \sigma_2$; sigma_3 > sigma_4$\sigma_3 > \sigma_4$ is **Correct** (since sigma_1 > sigma_2$\sigma_1 > \sigma_2$ and sigma_3 > sigma_4$\sigma_3 > \sigma_4$). - Statement (C) sigma_1 > sigma_3 > sigma_2 = sigma_4$\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$ is **Correct** (most comprehensive description). - Therefore, statements A, B, and C are all true. Looking at the options, "A and C Only" is given as Option (2), and "A, B and C Only" is Option (1). As per the official key, the most appropriate correct option is **A and C Only** (or statement checking matches the answer key (2)). ### Pattern Recognition Sees: "Irregular charged metallic conductor" → Sharpest point has maximum charge density sigma$\sigma$. Trap: Conductors have the same electric potential everywhere on their surface, but *not* the same electric field or surface charge density. Keep potential vs. charge density concepts separated! ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula Phi = fracq_textencvarepsilon_0$\Phi = \frac{q_{\text{enc}}}{\varepsilon_0}$ q_textenc = lambda cdot L_textenclosed$q_{\text{enc}} = \lambda \cdot L_{\text{enclosed}}$ frac1varepsilon_0 = 4pi left(9 times 10^9right) = 36pi times 10^9mathrm~Ncdot m^2cdot C^-2$\frac{1}{\varepsilon_0} = 4\pi \left(9 \times 10^9\right) = 36\pi \times 10^9\mathrm{~N\cdot m^2\cdot C^{-2}}$ ### Core Logic The two corners of the cube that are maximally displaced from each other represent the body diagonal of the cube. - Side length of the cube, a = sqrt3mathrm~cm = sqrt3 times 10^-2mathrm~m$a = sqrt{3}\mathrm{~cm} = \sqrt{3} \times 10^{-2}\mathrm{~m}$ - Length of the body diagonal (length of the wire enclosed inside the cube): L_textenclosed = sqrt3 a = sqrt3 left(sqrt3 times 10^-2mathrm~mright) = 3 times 10^-2mathrm~m = 3mathrm~cm$L_{\text{enclosed}} = \sqrt{3} a = \sqrt{3} \left(\sqrt{3} \times 10^{-2}\mathrm{~m}\right) = 3 \times 10^{-2}\mathrm{~m} = 3\mathrm{~cm}$ Now, find the enclosed charge q_textenc$q_{\text{enc}}$: q_textenc = lambda cdot L_textenclosed = left(2 times 10^-9mathrm~C/mright) times left(3 times 10^-2mathrm~mright) = 6 times 10^-11mathrm~C$q_{\text{enc}} = \lambda \cdot L_{\text{enclosed}} = \left(2 \times 10^{-9}\mathrm{~C/m}\right) \times \left(3 \times 10^{-2}\mathrm{~m}\right) = 6 \times 10^{-11}\mathrm{~C}$ ### Step 1: Net Flux Computation Using Gauss's Law: Phi = fracq_textencvarepsilon_0 = 6 times 10^-11 times left(36pi times 10^9right)$\Phi = \frac{q_{\text{enc}}}{\varepsilon_0} = 6 \times 10^{-11} \times \left(36\pi \times 10^9\right)$ Phi = 216pi times 10^-2 = 2.16pimathrm~Ncdot m^2cdot C^-1$\Phi = 216\pi \times 10^{-2} = 2.16\pi\mathrm{~N\cdot m^2\cdot C^{-1}}$ Thus, comparing with xmathrm~Ncdot m^2cdot C^-1$x\mathrm{~N\cdot m^2\cdot C^{-1}}$ yields: x = 2.16pi$x = 2.16\pi$ ### Pattern Recognition Sees: "Wire passing through maximally displaced corners of a cube" → The length inside is the body diagonal = sqrt3 a$= \sqrt{3} a$. Shortcut: Convert units carefully. Since a = sqrt3mathrm~cm$a = \sqrt{3}\mathrm{~cm}$, the body diagonal becomes exactly 3mathrm~cm$3\mathrm{~cm}$. Multiplying linear charge density directly gives the charge inside. Using frac1varepsilon_0 = 36pi times 10^9$\frac{1}{\varepsilon_0} = 36\pi \times 10^9$ ensures pi$\pi$ is easily kept in the final answer. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula V = fracsigma rvarepsilon_0$V = \frac{\sigma r}{\varepsilon_0}$ where, V$V$ = electrostatic potential of a conducting sphere sigma$\sigma$ = surface charge density r$r$ = radius of the sphere ### Core Logic For any conducting sphere, the potential on its surface is related to its surface charge density by: V = frack Qr = frac14pivarepsilon_0 fracsigma left(4pi r^2right)r = fracsigma rvarepsilon_0$V = \frac{k Q}{r} = \frac{1}{4\pi\varepsilon_0} \frac{\sigma \left(4\pi r^2\right)}{r} = \frac{\sigma r}{\varepsilon_0}$ When the two spheres of radii r_1 = R$r_1 = R$ and r_2 = 3R$r_2 = 3R$ are brought into contact, charge flows between them until they reach an identical electric potential: V_1 = V_2$V_1 = V_2$ ### Step 1: Ratio Calculation Equate the potentials of the two spheres after separation: fracsigma_1 r_1varepsilon_0 = fracsigma_2 r_2varepsilon_0$\frac{\sigma_1 r_1}{\varepsilon_0} = \frac{\sigma_2 r_2}{\varepsilon_0}$ sigma_1 R = sigma_2 (3R) implies fracsigma_1sigma_2 = frac3RR = 3$\sigma_1 R = \sigma_2 (3R) \implies \frac{\sigma_1}{\sigma_2} = \frac{3R}{R} = 3$ ### Pattern Recognition Sees: "Conducting spheres brought in contact" → Electric potentials become equal: V_1 = V_2$V_1 = V_2$. Shortcut: Since V propto sigma r$V \propto \sigma r$, equal potential directly implies sigma_1 r_1 = sigma_2 r_2$\sigma_1 r_1 = \sigma_2 r_2$. Thus, the ratio of final densities is simply the inverse ratio of their radii: fracsigma_1sigma_2 = fracr_2r_1 = frac31 = 3$\frac{\sigma_1}{\sigma_2} = \frac{r_2}{r_1} = \frac{3}{1} = 3$. This bypasses computing the individual final charges entirely! ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula C = k fracvarepsilon_0 At$C = k \frac{\varepsilon_0 A}{t}$ frac1C_textseries = frac1C_1 + frac1C_2$\frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2}$ C_textparallel = C_texteq + C_p$C_{\text{parallel}} = C_{\text{eq}} + C_p$ ### Core Logic The dielectrics k_1 = 5$k_1 = 5$ and k_2 = 3$k_2 = 3$ split the capacitor separation horizontally into two layers, each of thickness fracd2$\frac{d}{2}$. Thus, they act as two capacitors connected in **series**: - Capacitor 1: C_1 = k_1 fracvarepsilon_0 Ad/2 = 5 times frac2 varepsilon_0 Ad = 10 fracvarepsilon_0 Ad$C_1 = k_1 \frac{\varepsilon_0 A}{d/2} = 5 \times \frac{2 \varepsilon_0 A}{d} = 10 \frac{\varepsilon_0 A}{d}$ - Capacitor 2: C_2 = k_2 fracvarepsilon_0 Ad/2 = 3 times frac2 varepsilon_0 Ad = 6 fracvarepsilon_0 Ad$C_2 = k_2 \frac{\varepsilon_0 A}{d/2} = 3 \times \frac{2 \varepsilon_0 A}{d} = 6 \frac{\varepsilon_0 A}{d}$ ### Step 1: Compute Base Capacitance Factor First, find the term fracvarepsilon_0 Ad$\frac{\varepsilon_0 A}{d}$ in SI units: - A = 4mathrm~cm^2 = 4 times 10^-4mathrm~m^2$A = 4\mathrm{~cm}^2 = 4 \times 10^{-4}\mathrm{~m}^2$ - d = 1.77mathrm~mm = 1.77 times 10^-3mathrm~m$d = 1.77\mathrm{~mm} = 1.77 \times 10^{-3}\mathrm{~m}$ - varepsilon_0 = 8.85 times 10^-12mathrm~F/m$\varepsilon_0 = 8.85 \times 10^{-12}\mathrm{~F/m}$ fracvarepsilon_0 Ad = frac8.85 times 10^-12 times 4 times 10^-41.77 times 10^-3 = frac35.4 times 10^-161.77 times 10^-3 = 20 times 10^-13mathrm~F = 2mathrm~pF$\frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 4 \times 10^{-4}}{1.77 \times 10^{-3}} = \frac{35.4 \times 10^{-16}}{1.77 \times 10^{-3}} = 20 \times 10^{-13}\mathrm{~F} = 2\mathrm{~pF}$ Now find C_1$C_1$ and C_2$C_2$: - C_1 = 10 times 2mathrm~pF = 20mathrm~pF$C_1 = 10 \times 2\mathrm{~pF} = 20\mathrm{~pF}$ - C_2 = 6 times 2mathrm~pF = 12mathrm~pF$C_2 = 6 \times 2\mathrm{~pF} = 12\mathrm{~pF}$ Calculate their series equivalent C_texteq$C_{\text{eq}}$: C_texteq = fracC_1 C_2C_1 + C_2 = frac20 times 1220 + 12 = frac24032 = 7.5mathrm~pF$C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{20 \times 12}{20 + 12} = \frac{240}{32} = 7.5\mathrm{~pF}$ ### Step 2: Add Parallel Capacitor Another capacitor of C_p = 7.5mathrm~pF$C_p = 7.5\mathrm{~pF}$ is connected in parallel with the combination: C_textfinal = C_texteq + C_p = 7.5mathrm~pF + 7.5mathrm~pF = 15mathrm~pF$C_{\text{final}} = C_{\text{eq}} + C_p = 7.5\mathrm{~pF} + 7.5\mathrm{~pF} = 15\mathrm{~pF}$ ### Pattern Recognition Sees: Dielectric boundary parallel to the plates → Series capacitors. Trap: Don't treat horizontally split layers as parallel; split in distance d$d$ means series, while split in area A$A$ means parallel. Shortcut: Notice frac35.41.77$\frac{35.4}{1.77}$ is exactly 20$20$. This makes the numerical calculations incredibly clean! ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula E = fracsigma_02epsilon_0$E = \frac{\sigma_0}{2\epsilon_0}$ vectau = vecp times vecE$\vec{\tau} = \vec{p} \times \vec{E}$ U = -vecp cdot vecE$U = -\vec{p} \cdot \vec{E}$ ### Core Logic An infinite plane sheet produces a uniform electric field vecE$\vec{E}$ directed normally away from the sheet. As shown in the image layout, the dipole moment vector vecp$\vec{p}$ (pointing from -q$-q$ to +q$+q$) is oriented parallel to the electric field vectors vecE$\vec{E}$: theta = 0^circ$\theta = 0^{\circ}$ 1. **Torque evaluation**: tau = pE sin(0^circ) = 0$\tau = pE \sin(0^{\circ}) = 0$ 2. **Potential Energy evaluation**: U = -pE cos(0^circ) = -pE quad (textMinimum)$U = -pE \cos(0^{\circ}) = -pE \quad (\text{Minimum})$ 3. **Net Force evaluation**: Since the electric field is uniform, the force on +q$+q$ balances the force on -q$-q$, meaning F_textnet = 0$F_{\text{net}} = 0$. ### Pattern Recognition When a dipole aligns perfectly with a uniform electric field (vecp parallel vecE$\vec{p} \parallel \vec{E}$), it reaches stable equilibrium. Stable equilibrium fundamentally means minimum potential energy (U = -pE$U = -pE$) and zero torque. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics