A metallic ring is uniformly charged as shown in figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc AB to 'O' is 'E' is magnitude. What would be the magnitude of electric field at 'O' due to arc ABC?

### 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

### 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