### 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

### Related Formula The electrostatic energy density u$u$ stored in an electric field E$E$ is given by: u = frac12epsilon_0E^2$u = \frac{1}{2}\epsilon_{0}E^{2}$ The total potential energy U$U$ stored in a volume V$V$ is: U = u cdot V$U = u \cdot V$ ### Core Logic For a parallel plate capacitor, the volume between the plates where the electric field exists is the product of the plate area A$A$ and the plate separation d$d$: V = Ad$V = Ad$ ### Step 1: Calculating Stored Energy Substitute the volume expression into the total energy equation: U = left(frac12epsilon_0E^2 ight)(Ad)$U = \left(\frac{1}{2}\epsilon_{0}E^{2} ight)(Ad)$ U = frac12epsilon_0E^2Ad$U = \frac{1}{2}\epsilon_{0}E^{2}Ad$ ### Pattern Recognition Energy density times volume is a universal relation for field fields. Remember that volume is simply cross-sectional area multiplied by distance (Ad$Ad$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula The capacitance of a parallel plate system is given by : C = fracepsilon_0Ad = fracepsilon_0lbd$C = \frac{\epsilon_{0}A}{d} = \frac{\epsilon_{0}lb}{d}$ where l$l$ is length, b$b$ is breadth, and d$d$ is separation distance. ### Core Logic Evaluate the initial capacitance base scaling parameter : C_0 = fracepsilon_0 times 3text cm times 1text cm3mutextm = 1 times epsilon_0text units$C_{0} = \frac{\epsilon_{0} \times 3\text{ cm} \times 1\text{ cm}}{3\mu\text{m}} = 1 \times \epsilon_{0}\text{ units}$ We want to increase this initial baseline capacitance value by a factor of 10, meaning our target capacitance is 10epsilon_0$10\epsilon_{0}$. ### Step 1: Audit Options Let's check the capacitance for options C and E : * Option C: l=6text cm, b=5text cm, d=3mutextm$l=6\text{ cm}, b=5\text{ cm}, d=3\mu\text{m}$ . C_C = fracepsilon_0 times 6 times 53 = 10epsilon_0text units (Correct)$C_{C} = \frac{\epsilon_{0} \times 6 \times 5}{3} = 10\epsilon_{0}\text{ units (Correct)}$ * Option E: l=5text cm, b=2text cm, d=1mutextm$l=5\text{ cm}, b=2\text{ cm}, d=1\mu\text{m}$ . C_E = fracepsilon_0 times 5 times 21 = 10epsilon_0text units (Correct)$C_{E} = \frac{\epsilon_{0} \times 5 \times 2}{1} = 10\epsilon_{0}\text{ units (Correct)}$ ### Pattern Recognition Capacitance scales matching the geometric factor fracl cdot bd$\frac{l \cdot b}{d}$. Look for options where this ratio scales up to exactly 10 times the initial baseline value. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics