### Related Formula phi = fracq_textenclosedepsilon_0$\phi = \frac{q_{\text{enclosed}}}{\epsilon_0}$ where, phi$\phi$ = net electric flux through the closed surface q_textenclosed$q_{\text{enclosed}}$ = net charge enclosed by the surface epsilon_0$\epsilon_0$ = permittivity of free space ### Core Logic According to Gauss's Law, the total electric flux through a closed surface depends only on the charge enclosed inside it, completely independent of the radius of the surface. Rearranging the formula to solve for q$q$: q = phi cdot epsilon_0$q = \phi \cdot \epsilon_0$ Substitute the given values: q = (-2 times 10^4 mathrm~Nm^2mathrmC^-1) times (8.85 times 10^-12 mathrm~C^2mathrmN^-1mathrmm^-2)$q = (-2 \times 10^{4} \mathrm{~Nm}^2\mathrm{C}^{-1}) \times (8.85 \times 10^{-12} \mathrm{~C}^2\mathrm{N}^{-1}\mathrm{m}^{-2})$ q = -17.7 times 10^-8 mathrm~C$q = -17.7 \times 10^{-8} \mathrm{~C}$ ### Pattern Recognition Distractor alert: The radius (8.0mathrm~cm$8.0\mathrm{~cm}$) is extra data meant to mislead. Gauss's flux depends entirely on the magnitude of the enclosed charge, not the physical surface area configuration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula Capacitance with dielectric: C = fracvarepsilon_r varepsilon_0 Ad$C = \frac{\varepsilon_r \varepsilon_0 A}{d}$ Series Capacitors: C_texteq = fracC_a C_bC_a + C_b$C_{\text{eq}} = \frac{C_a C_b}{C_a + C_b}$ Parallel Capacitors: C_texteq = C_a + C_b$C_{\text{eq}} = C_a + C_b$ ### Core Logic Let C_0 = fracvarepsilon_0 Ad$C_0 = \frac{\varepsilon_0 A}{d}$ be the capacitance without any dielectric. - **First Configuration (Series connection)**: The dielectrics split the gap vertically, so the effective thickness of each slab is d/2$d/2$, and the area remains A$A$. C_a = fracvarepsilon_1 varepsilon_0 Ad/2 = 2varepsilon_1 C_0$C_a = \frac{\varepsilon_1 \varepsilon_0 A}{d/2} = 2\varepsilon_1 C_0$ C_b = fracvarepsilon_2 varepsilon_0 Ad/2 = 2varepsilon_2 C_0$C_b = \frac{\varepsilon_2 \varepsilon_0 A}{d/2} = 2\varepsilon_2 C_0$ Since they are in series: C_1 = fracC_a C_bC_a + C_b = frac(2varepsilon_1 C_0)(2varepsilon_2 C_0)2varepsilon_1 C_0 + 2varepsilon_2 C_0 = frac4varepsilon_1varepsilon_2 C_0^22C_0(varepsilon_1 + varepsilon_2) = frac2varepsilon_1varepsilon_2varepsilon_1 + varepsilon_2 C_0$C_1 = \frac{C_a C_b}{C_a + C_b} = \frac{(2\varepsilon_1 C_0)(2\varepsilon_2 C_0)}{2\varepsilon_1 C_0 + 2\varepsilon_2 C_0} = \frac{4\varepsilon_1\varepsilon_2 C_0^2}{2C_0(\varepsilon_1 + \varepsilon_2)} = \frac{2\varepsilon_1\varepsilon_2}{\varepsilon_1 + \varepsilon_2} C_0$ - **Second Configuration (Parallel connection)**: The dielectrics split the area horizontally, so the effective area of each slab is A/2$A/2$, and the distance remains d$d$. C_c = fracvarepsilon_1 varepsilon_0 (A/2)d = fracvarepsilon_1 C_02$C_c = \frac{\varepsilon_1 \varepsilon_0 (A/2)}{d} = \frac{\varepsilon_1 C_0}{2}$ C_d = fracvarepsilon_2 varepsilon_0 (A/2)d = fracvarepsilon_2 C_02$C_d = \frac{\varepsilon_2 \varepsilon_0 (A/2)}{d} = \frac{\varepsilon_2 C_0}{2}$ Since they are in parallel: C_2 = C_c + C_d = (varepsilon_1 + varepsilon_2) fracC_02$C_2 = C_c + C_d = (\varepsilon_1 + \varepsilon_2) \frac{C_0}{2}$

### Step 1: Calculating the Ratio Now, compute fracC_1C_2$\frac{C_1}{C_2}$: fracC_1C_2 = fracleft(frac2varepsilon_1varepsilon_2varepsilon_1 + varepsilon_2right) C_0left(fracvarepsilon_1 + varepsilon_22right) C_0 = frac4varepsilon_1varepsilon_2(varepsilon_1 + varepsilon_2)^2$\frac{C_1}{C_2} = \frac{\left(\frac{2\varepsilon_1\varepsilon_2}{\varepsilon_1 + \varepsilon_2}\right) C_0}{\left(\frac{\varepsilon_1 + \varepsilon_2}{2}\right) C_0} = \frac{4\varepsilon_1\varepsilon_2}{(\varepsilon_1 + \varepsilon_2)^2}$ ### Pattern Recognition For dielectric-filled capacitors: splitting the gap (d/2$d/2$) leads to a series combination, while splitting the plate area (A/2$A/2$) leads to a parallel combination. Shortcut: C_textseries = textharmonic mean$C_{\text{series}} = \text{harmonic mean}$, C_textparallel = textarithmetic mean$C_{\text{parallel}} = \text{arithmetic mean}$. The ratio fracC_1C_2$\frac{C_1}{C_2}$ is always the ratio of the harmonic mean of the dielectric constants to their arithmetic mean! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance

### Related Formula For a uniformly charged spherical shell of radius R$R$ and charge Q$Q$: - Inside and on the surface of the shell (r le R$r \le R$): V_textin = V_textsurface = frackQR$V_{\text{in}} = V_{\text{surface}} = \frac{kQ}{R}$ - Outside the shell (r > R$r > R$): V_textout = frackQr = V_textsurface left(fracRrright)$V_{\text{out}} = \frac{kQ}{r} = V_{\text{surface}} \left(\frac{R}{r}\right)$ ### Core Logic Let's calculate the potentials at the specified positions: - Given surface potential at R = 10mathrm~cm$R = 10\mathrm{~cm}$ is 120mathrm~V$120\mathrm{~V}$. 1. **At the center (r = 0$r = 0$)**: Since the center lies inside the shell (0 < 10mathrm~cm$0 < 10\mathrm{~cm}$), the potential equals the surface potential: V_textcentre = 120mathrm~V$V_{\text{centre}} = 120\mathrm{~V}$ 2. **At r = 5mathrm~cm$r = 5\mathrm{~cm}$**: Since 5mathrm~cm$5\mathrm{~cm}$ is also inside the shell (5 < 10mathrm~cm$5 < 10\mathrm{~cm}$), the potential remains constant at the surface value: V_r=5 = 120mathrm~V$V_{r=5} = 120\mathrm{~V}$ 3. **At r = 15mathrm~cm$r = 15\mathrm{~cm}$**: Since 15mathrm~cm$15\mathrm{~cm}$ is outside the shell (15 > 10mathrm~cm$15 > 10\mathrm{~cm}$), the potential decreases inversely with distance: V_r=15 = V_textsurface left(fracRrright) = 120 times frac1015 = 80mathrm~V$V_{r=15} = V_{\text{surface}} \left(\frac{R}{r}\right) = 120 \times \frac{10}{15} = 80\mathrm{~V}$ Therefore, the potentials are 120mathrm~V$120\mathrm{~V}$, 120mathrm~V$120\mathrm{~V}$, and 80mathrm~V$80\mathrm{~V}$ respectively. ### Pattern Recognition The electric field inside a uniformly charged conducting spherical shell is zero, meaning that no work is done moving a charge inside it. Consequently, the potential remains absolutely uniform/constant from the surface all the way to the center! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance

### Related Formula The electric field due to a circular arc subtending an angle phi$\phi$ at the center is given by: E_textarc = frac2klambdaR sinleft(fracphi2right)$E_{\text{arc}} = \frac{2k\lambda}{R} \sin\left(\frac{\phi}{2}\right)$ ### Core Logic Arc AB subtends 90^circ$90^\circ$ (one quadrant) at the center. The electric field due to it is given as E$E$. Arc ABC consists of two independent quadrants: arc AB and arc BC. Each quadrant independently creates an electric field of magnitude E$E$ pointing along the bisector of that specific quadrant. ### Step 1: Vector Addition The electric field vecE_AB$\vec{E}_{AB}$ is directed at 45^circ$45^\circ$ away from both axes into the third quadrant. The electric field vecE_BC$\vec{E}_{BC}$ is directed at 45^circ$45^\circ$ towards the matching opposite quadrant. Since vecE_AB$\vec{E}_{AB}$ and vecE_BC$\vec{E}_{BC}$ are perpendicular to each other, their resultant magnitude is: E_textnet = sqrtE^2 + E^2 = sqrt2E$E_{\text{net}} = \sqrt{E^2 + E^2} = \sqrt{2}E$

### Pattern Recognition Symmetric components of a ring create orthogonal vector fields. Each 90^circ$90^\circ$ arc produces a field of magnitude E$E$ directed along its angular bisector. Two adjacent quadrants have bisectors separated by 90^circ$90^\circ$, hence use orthogonal vector addition. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula Capacitance modification by dielectric: C' = K cdot C$C' = K \cdot C$ Series combination: C_textseries = fracC_a C_bC_a + C_b$C_{\text{series}} = \frac{C_a C_b}{C_a + C_b}$ Parallel combination: C_textparallel = C_1 + C_2$C_{\text{parallel}} = C_1 + C_2$ ### Core Logic Initial capacitance value C = 5\ mutextF$C = 5\ \mu\text{F}$ for all. After dielectric insertion into C_1$C_1$, its value becomes: C_1 = 4 times 5 = 20\ mutextF$C_1 = 4 \times 5 = 20\ \mu\text{F}$ The values for the others remain constant: C_2 = 5\ mutextF, quad C_3 = 5\ mutextF$C_2 = 5\ \mu\text{F}, \quad C_3 = 5\ \mu\text{F}$ ### Step 1: Circuit Topology Analysis From the network layout, C_1$C_1$ and C_2$C_2$ are configured in a series arm, which is collectively in parallel with C_3$C_3$. Equivalent of the series arm: C_12 = frac20 times 520 + 5 = frac10025 = 4\ mutextF$C_{12} = \frac{20 \times 5}{20 + 5} = \frac{100}{25} = 4\ \mu\text{F}$ Adding the parallel branch C_3$C_3$: C_texteq = C_12 + C_3 = 4 + 5 = 9\ mutextF$C_{\text{eq}} = C_{12} + C_3 = 4 + 5 = 9\ \mu\text{F}$ ### Pattern Recognition Identify layout components systematically. Series components simplify via product-over-sum, then combine linearly with parallel components. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics