A point charge +q$+q$ is placed at the origin. A second point charge +9q$+9q$ is placed at (d, 0, 0)$(d, 0, 0)$ in Cartesian coordinate system. The point in between them where the electric field vanishes is:

### 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