Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): Net dipole moment of a polar linear isotropic dielectric substance is not zero even in the absence of an external electric field. Reason (R): In absence of an external electric field, the different permanent dipoles of a polar dielectric substance are oriented in random directions. In the light of the above statements, choose the most appropriate answer from the options given below:

### Core Logic According to electrostatic shielding, the electric field inside a cavity of a conductor is always zero, regardless of the size and shape of the cavity and regardless of any charges located outside or on the conductor's surface[cite: 660]. Therefore, when lightning strikes a metal aircraft, the entire charge stays on the outer metallic surface and flows down without producing an electric field inside, keeping passengers safe[cite: 12]. ### Step 1: Statement Evaluation * **Assertion (A):** Correct, passengers are protected from lightning because of the metallic body shield [cite: 12]. * **Reason (R):** Correct, the field inside a cavity of a conductor is zero [cite: 13]. * **Explanation:** Since the zero field property is precisely why passengers are protected, (R) correctly explains (A)[cite: 15]. ### Pattern Recognition Metallic shell / shield configuration always establishes E_textinside = 0$E_{\text{inside}} = 0$[cite: 660]. This shielding mechanism directly underpins safety features in lightning scenarios for cars and airplanes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula E = fracVd$$E = \frac{V}{d}$$ [cite: 736] tau = pEsintheta$$\tau = pE\sin\theta$$ [cite: 737] p = q cdot a$p = q \cdot a$ [cite: 738] ### Core Logic First, calculate the electric field magnitude E$E$ between the capacitor plates: [cite: 58, 736] E = frac50.5 times 10^-3 = 10^4\ textV/m$$E = \frac{5}{0.5 \times 10^{-3}} = 10^4\ \text{V/m}$$ [cite: 58, 59, 736] Next, evaluate the dipole moment p$p$: [cite: 58, 738] p = (2 times 10^-6\ textC) times (0.5 times 10^-6\ textm) = 1 times 10^-12\ textCcdottextm$$p = (2 \times 10^{-6}\ \text{C}) \times (0.5 \times 10^{-6}\ \text{m}) = 1 \times 10^{-12}\ \text{C}\cdot\text{m}$$ [cite: 58, 740] Now find the torque when rotated by theta = 30^circ$\theta = 30^{\circ}$: [cite: 59, 737] tau = (1 times 10^-12) times 10^4 times sin 30^circ = 10^-8 times frac12 = 5 times 10^-9\ textNcdottextm$$\tau = (1 \times 10^{-12}) \times 10^4 \times \sin 30^{\circ} = 10^{-8} \times \frac{1}{2} = 5 \times 10^{-9}\ \text{N}\cdot\text{m}$$ [cite: 743] ### Pattern Recognition Always convert parameters to pristine standard SI units before applying electrostatic expressions (0.5\ mutextm = 5 times 10^-7\ textm$0.5\ \mu\text{m} = 5 \times 10^{-7}\ \text{m}$ and 0.5\ textmm = 5 times 10^-4\ textm$0.5\ \text{mm} = 5 \times 10^{-4}\ \text{m}$) to secure zero conversion error[cite: 58, 59, 736, 738]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula Q_textind = Qleft(1 - frac1Kright)$$Q_{\text{ind}} = Q\left(1 - \frac{1}{K}\right)$$ [cite: 813] ### Core Logic Substitute the values given for free surface charge Q = 5 times 10^-6\ textC$Q = 5 \times 10^{-6}\ \text{C}$ and bound induced charge Q_textind = 4 times 10^-6\ textC$Q_{\text{ind}} = 4 \times 10^{-6}\ \text{C}$ into the equation: [cite: 183, 184, 814] 4 times 10^-6 = 5 times 10^-6 left(1 - frac1Kright)$$4 \times 10^{-6} = 5 \times 10^{-6} \left(1 - \frac{1}{K}\right)$$ [cite: 814] frac45 = 1 - frac1K implies frac1K = 1 - frac45 = frac15$$\frac{4}{5} = 1 - \frac{1}{K} \implies \frac{1}{K} = 1 - \frac{4}{5} = \frac{1}{5}$$ [cite: 815] K = 5$K = 5$ [cite: 815] ### Pattern Recognition The fraction of charge induced on the dielectric face scales structurally as fracK-1K$\frac{K-1}{K}$[cite: 813, 815]. Observing a ratio of 4$4$ parts out of 5$5$ implies that the constant factor K$K$ must equal 5$5$ directly[cite: 815]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula phi = vecE cdot vecA$$\phi = \vec{E} \cdot \vec{A}$$ [cite: 827] ### Core Logic A surface aligned parallel to the xtext-z$x\text{-}z$ plane possesses an area vector pointing completely orthogonal to it along the y$y$-axis direction, meaning vecA = Ahatj$\vec{A} = A\hat{j}$[cite: 196, 827]. Performing the dot product: [cite: 827] phi = left[(2hati + 4hatj + 6hatk) times 10^3right] cdot (Ahatj) = 4 times 10^3 A$$\phi = \left[(2\hat{i} + 4\hat{j} + 6\hat{k}) \times 10^3\right] \cdot (A\hat{j}) = 4 \times 10^3 A$$ [cite: 195, 827] Given that the net flux magnitude is 6.0\ textNm^2textC^-1$6.0\ \text{Nm}^2\text{C}^{-1}$ [cite: 196]: 6 = 4 times 10^3 A implies A = frac64 times 10^3 = 1.5 times 10^-3\ textm^2$$6 = 4 \times 10^3 A \implies A = \frac{6}{4 \times 10^3} = 1.5 \times 10^{-3}\ \text{m}^2$$ [cite: 828, 829] Converting square meters to square centimeters (1\ textm^2 = 10^4\ textcm^2$1\ \text{m}^2 = 10^4\ \text{cm}^2$): [cite: 196, 830] A = 1.5 times 10^-3 times 10^4 = 15\ textcm^2$$A = 1.5 \times 10^{-3} \times 10^4 = 15\ \text{cm}^2$$ [cite: 830] ### Pattern Recognition Always focus exclusively on the specific field component matched to the surface orientation normal[cite: 827]. For an xtext-z$x\text{-}z$ plane match, only the hatj$\hat{j}$ coefficient creates flux[cite: 196, 827]. Do not miss the metric scale unit transition at the end (m^2 rightarrow cm^2$m^2 \rightarrow cm^2$)[cite: 196, 830]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula F = frack q_1 q_2r^2$$F = \frac{k q_1 q_2}{r^2}$$ ### Core Logic When two identical conducting spheres are brought into contact, the total initial charge splits equally between them: q_1 = q_2 = frac4 times 10^-8\ mathrmC + 02 = 2 times 10^-8\ mathrmC$$q_1 = q_2 = \frac{4 \times 10^{-8}\ \mathrm{C} + 0}{2} = 2 \times 10^{-8}\ \mathrm{C}$$ Given repulsion force, F = 9 times 10^-3\ mathrmN$F = 9 \times 10^{-3}\ \mathrm{N}$: 9 times 10^-3 = frac9 times 10^9 times (2 times 10^-8) times (2 times 10^-8)r^2$$9 \times 10^{-3} = \frac{9 \times 10^{9} \times (2 \times 10^{-8}) \times (2 \times 10^{-8})}{r^2}$$ 9 times 10^-3 = frac36 times 10^-7r^2 implies r^2 = frac36 times 10^-79 times 10^-3 = 4 times 10^-4$$9 \times 10^{-3} = \frac{36 \times 10^{-7}}{r^2} \implies r^2 = \frac{36 \times 10^{-7}}{9 \times 10^{-3}} = 4 \times 10^{-4}$$ r = 2 times 10^-2\ mathrmm = 2\ mathrmcm$$r = 2 \times 10^{-2}\ \mathrm{m} = 2\ \mathrm{cm}$$

### Pattern Recognition Identical spheres in contact distribute net charge equally due to symmetric capacitance sharing: q' = Q_texttotal / 2$q' = Q_{\text{total}} / 2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics