Electric field in a region is given by vecE = Axhati + Byhatj$\vec{E} = Ax\hat{i} + By\hat{j}$, where A = 10~mathrmV / m^2$A = 10~\mathrm{V / m^2}$ and B = 5~mathrmV / m^2$B = 5~\mathrm{V / m^2}$. If the electric potential at a point (10, 20) is 500~mathrmV$500~\mathrm{V}$, then the electric potential at origin is \_\_\_\_ V.

### Related Formula C = fracepsilon_0 Ad - t + fractK$$C = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}$$ Alternatively, treat as two capacitors in series: C_eq = fracC_1 C_2C_1 + C_2$C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$ ### Core Logic Initial capacitance (vacuum): C = fracA epsilon_0d$C = \frac{A \epsilon_{0}}{d}$.

The system can be modelled as two capacitors in series: 1. A vacuum capacitor of thickness d - fracd3 = frac2d3$d - \frac{d}{3} = \frac{2d}{3}$ 2. A dielectric capacitor of thickness fracd3$\frac{d}{3}$ and dielectric constant K$K$.

### Step 1: Capacitors in Series The capacitances are: C_1 = fracepsilon_0 Aleft(frac2d3right) = frac32 left(fracepsilon_0 Adright) = frac32 C$$C_1 = \frac{\epsilon_0 A}{\left(\frac{2d}{3}\right)} = \frac{3}{2} \left(\frac{\epsilon_0 A}{d}\right) = \frac{3}{2} C$$ C_2 = fracK epsilon_0 Aleft(fracd3right) = 3K left(fracepsilon_0 Adright) = 3KC$$C_2 = \frac{K \epsilon_0 A}{\left(\frac{d}{3}\right)} = 3K \left(\frac{\epsilon_0 A}{d}\right) = 3KC$$ Now, equivalent capacitance in series: C_texteq = fracC_1 C_2C_1 + C_2 = fracleft(frac32 Cright) times (3KC)frac32 C + 3KC$$C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{\left(\frac{3}{2} C\right) \times (3KC)}{\frac{3}{2} C + 3KC}$$ C_texteq = fracfrac92KC^2frac32C(1 + 2K)$$C_{\text{eq}} = \frac{\frac{9}{2}KC^2}{\frac{3}{2}C(1 + 2K)}$$ C_texteq = frac3KC2K + 1$$C_{\text{eq}} = \frac{3KC}{2K + 1}$$ ### Pattern Recognition Partial dielectric filling of thickness t$t$: Use formula C_textnew = fracepsilon_0 Ad - t + t/K$C_{\text{new}} = \frac{\epsilon_0 A}{d - t + t/K}$. Plugging t = d/3$t = d/3$ directly gives fracepsilon_0 Ad - d/3 + d/3K = fracepsilon_0 Afrac2d3 + fracd3K = frac3K epsilon_0 A2Kd + d = frac3KC2K+1$\frac{\epsilon_0 A}{d - d/3 + d/3K} = \frac{\epsilon_0 A}{\frac{2d}{3} + \frac{d}{3K}} = \frac{3K \epsilon_0 A}{2Kd + d} = \frac{3KC}{2K+1}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula W_textext = Delta U = U_f - U_i$$W_{\text{ext}} = \Delta U = U_f - U_i$$ U = frac14piepsilon_0 fracq_1 q_2r$$U = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$$ ### Core Logic Work done by external agent: W_textext = Delta U$W_{\text{ext}} = \Delta U$, where Delta U$\Delta U$ is the change in potential energy. W_textext = frac14pi epsilon_0 fracq_1 q_2r_f - frac14pi epsilon_0 fracq_1 q_2r_i$$W_{\text{ext}} = \frac{1}{4\pi \epsilon_{0}} \frac{q_{1} q_{2}}{r_{f}} - \frac{1}{4\pi \epsilon_{0}} \frac{q_{1} q_{2}}{r_{i}}$$ Calculate the distances of points A and B from the origin: r_i = |A| = sqrt4^2 + 4^2 + 2^2 = sqrt16+16+4 = sqrt36 = 6text m$r_i = |A| = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16+16+4} = \sqrt{36} = 6\text{ m}$ r_f = |B| = sqrt2^2 + 2^2 + 1^2 = sqrt4+4+1 = sqrt9 = 3text m$r_f = |B| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3\text{ m}$ ### Step 1: Calculate Work Done W_textext = (9 times 10^9) times (10^-8 times 2 times 10^-6) left[ frac13 - frac16 right]$$W_{\text{ext}} = (9 \times 10^9) \times (10^{-8} \times 2 \times 10^{-6}) \left[ \frac{1}{3} - \frac{1}{6} \right]$$ W_textext = 18 times 10^-5 times left(frac2-16right)$$W_{\text{ext}} = 18 \times 10^{-5} \times \left(\frac{2-1}{6}\right)$$ W_textext = 18 times 10^-5 times frac16 = 3 times 10^-5text J = 30 times 10^-6text J$$W_{\text{ext}} = 18 \times 10^{-5} \times \frac{1}{6} = 3 \times 10^{-5}\text{ J} = 30 \times 10^{-6}\text{ J}$$ ### Pattern Recognition Electric field is conservative. Work done simply equals change in kqq/r$kqq/r$ from initial to final radial coordinate. No path dependence. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula K_i + U_i = K_f + U_f$$K_i + U_i = K_f + U_f$$ U = frackq_1q_2r - fracGm_1m_2r$$U = \frac{kq_1q_2}{r} - \frac{Gm_1m_2}{r}$$ ### Core Logic Using energy conservation from the initial state (separation 4R$4R$) to the final state (just touching, so center-to-center separation is 2R$2R$). Both spheres have mass m$m$ and speed u$u$. Initial Energy: E_i = 2left(frac12mu^2right) - fracGm^24R + frackQ^24R$$E_i = 2\left(\frac{1}{2}mu^2\right) - \frac{Gm^2}{4R} + \frac{kQ^2}{4R}$$ Final Energy (just touching implies final velocity is zero): E_f = - fracGm^22R + frackQ^22R$$E_f = - \frac{Gm^2}{2R} + \frac{kQ^2}{2R}$$ ### Step 1: Equating Energies mu^2 - fracGm^24R + frackQ^24R = - fracGm^22R + frackQ^22R$$mu^2 - \frac{Gm^2}{4R} + \frac{kQ^2}{4R} = - \frac{Gm^2}{2R} + \frac{kQ^2}{2R}$$ mu^2 = left(frackQ^22R - frackQ^24Rright) - left(fracGm^22R - fracGm^24Rright)$$mu^2 = \left(\frac{kQ^2}{2R} - \frac{kQ^2}{4R}\right) - \left(\frac{Gm^2}{2R} - \frac{Gm^2}{4R}\right)$$ mu^2 = frackQ^24R - fracGm^24R$$mu^2 = \frac{kQ^2}{4R} - \frac{Gm^2}{4R}$$ u^2 = frac14mR(kQ^2 - Gm^2)$$u^2 = \frac{1}{4mR}(kQ^2 - Gm^2)$$ ### Step 2: Final Conclusion u = sqrtfrackQ^24mRleft(1 - fracGm^2kQ^2right)$$u = \sqrt{\frac{kQ^2}{4mR}\left(1 - \frac{Gm^2}{kQ^2}\right)}$$ ### Pattern Recognition When dealing with two forces (electrostatic repulsion and gravitational attraction), their potentials simply superimpose linearly. The change in total potential energy equals the loss in kinetic energy. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics Class 11 Physics: Gravitation

### Related Formula Q = C V_c$Q = C V_c$ where V_c$V_c$ is the steady-state voltage across the capacitor. ### Core Logic In steady state, the capacitor acts as an open circuit (blocks DC current). We must analyze the active branches.

The branch with the reversed-biased diode will carry no current. Current flows through the outer loop via the forward-biased diode. Total active resistance R_texteq = 1\,Omega + 4\,Omega = 5\,Omega$R_{\text{eq}} = 1\,\Omega + 4\,\Omega = 5\,\Omega$. Current in the steady state: i = fracVR_texteq = frac2.55 = 0.5 text A$$i = \frac{V}{R_{\text{eq}}} = \frac{2.5}{5} = 0.5 \text{ A}$$ ### Step 1: Voltage Calculation The voltage across the capacitor V_c$V_c$ is equal to the voltage drop across the 4\,Omega$4\,\Omega$ resistor, because the branch is connected in parallel. V_c = i times 4\,Omega$$V_c = i \times 4\,\Omega$$ V_c = 0.5 times 4 = 2 text V$$V_c = 0.5 \times 4 = 2 \text{ V}$$ ### Step 2: Final Conclusion The charge stored is: Q = C V_c$Q = C V_c$ Q = 5\,mutextF times 2text V = 10\,mutextC$$Q = 5\,\mu\text{F} \times 2\text{ V} = 10\,\mu\text{C}$$ ### Pattern Recognition Capacitor in steady-state DC = open wire. Analyze only the paths where current can physically flow, check diode polarity, find nodal voltage across the capacitor terminals, apply Q=CV. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics Class 12 Physics: Current Electricity Class 12 Physics: Semiconductor Electronics

### Related Formula E = frackQr^2$$E = \frac{kQ}{r^2}$$ ### Core Logic

By symmetry and vector addition of electric fields due to the point charges placed on the circle: vecE_textnet = 2E_0 cos 30^circ(-hati) + 2E_0 sin 30^circ(hatj)$$\vec{E}_{\text{net}} = 2E_0 \cos 30^{\circ}(-\hat{i}) + 2E_0 \sin 30^{\circ}(\hat{j})$$ = frac2kQr^2left[fracsqrt32(-hati) + frac12hatjright]$$= \frac{2kQ}{r^2}\left[\frac{\sqrt{3}}{2}(-\hat{i}) + \frac{1}{2}\hat{j}\right]$$ = -fracQ4piepsilon_0 R^2(sqrt3hati - hatj)$$= -\frac{Q}{4\pi\epsilon_0 R^2}(\sqrt{3}\hat{i} - \hat{j})$$ ### Pattern Recognition Sees: Symmetrical charge distribution on a circle. Shortcut: Resolve components symmetrically and combine vector contributions. Check: Matches option (2). ✓ ### Chapter Mix Class 12 Physics: Electrostatics