A parallel plate capacitor is filled equally (half) with two dielectrics of dielectric constant varepsilon_1$\varepsilon_1$ and varepsilon_2$\varepsilon_2$, as shown in figures. The distance between the plates is d$d$ and area of each plate is A$A$. If capacitance in first configuration and second configuration are C_1$C_1$ and C_2$C_2$ respectively, then fracC_1C_2$\frac{C_1}{C_2}$ is:

### 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 vecP_textnet = sum vecp_i$$\vec{P}_{\text{net}} = \sum \vec{p}_i$$ where: vecP_textnet$\vec{P}_{\text{net}}$ = net dipole moment of the dielectric vecp_i$\vec{p}_i$ = dipole moment of the individual i$i$-th molecule ### Core Logic No external electric field is present (E_textext = 0$E_{\text{ext}} = 0$). Due to thermal agitation, all molecular permanent dipoles are randomly oriented in space: vecP_textnet = 0 quad textwhen vecE_textext = 0$$\vec{P}_{\text{net}} = 0 \quad \text{when } \vec{E}_{\text{ext}} = 0$$ Thus: 1. Assertion (A) is false because it claims the net dipole moment is non-zero even without an external field. 2. Reason (R) is true because it correctly describes that different permanent dipoles are randomly oriented. ### Step 1: Final Conclusion Therefore, (A) is not correct but (R) is correct. ### Pattern Recognition Sees: "polar dielectric" + "no external field" → net bulk dipole moment is always zero. Trap: Confusing the molecular level with the macroscopic level. Each molecule in a polar dielectric has a permanent dipole moment, but the macro substance has zero net moment due to random thermal orientations. Shortcut: No external field means vectors cancel globally, which implies zero net moment. Thus (A) is false immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula E = fracsigma_textinnerepsilon_0$$E = \frac{\sigma_{\text{inner}}}{\epsilon_0}$$ ### Core Logic For parallel conducting plates of large area, charges redistribute on the outer and inner faces to maintain electrostatic equilibrium. Total charge per unit area on Plate 1: q_1 = sigma$q_1 = \sigma$ Total charge per unit area on Plate 2: q_2 = -2sigma$q_2 = -2\sigma$ The outer surface charge density on the far sides of both plates must be equal: sigma_textouter = fracq_1 + q_22 = fracsigma - 2sigma2 = -fracsigma2$$\sigma_{\text{outer}} = \frac{q_1 + q_2}{2} = \frac{\sigma - 2\sigma}{2} = -\frac{\sigma}{2}$$ Now, compute the charges on the inner facing surfaces: - Inner face of Plate 1: sigma_textinner1 = q_1 - sigma_textouter = sigma - left(-fracsigma2right) = frac3sigma2$$\sigma_{\text{inner1}} = q_1 - \sigma_{\text{outer}} = \sigma - \left(-\frac{\sigma}{2}\right) = \frac{3\sigma}{2}$$ - Inner face of Plate 2: sigma_textinner2 = q_2 - sigma_textouter = -2sigma - left(-fracsigma2right) = -frac3sigma2$$\sigma_{\text{inner2}} = q_2 - \sigma_{\text{outer}} = -2\sigma - \left(-\frac{\sigma}{2}\right) = -\frac{3\sigma}{2}$$ In the region between the plates, both inner surfaces create an electric field in the same direction (away from the positive plate 1 and towards negative plate 2): E = fracsigma_textinner12epsilon_0 + frac|sigma_textinner2|2epsilon_0 = frac3sigma/22epsilon_0 + frac3sigma/22epsilon_0 = frac3sigma2epsilon_0$$E = \frac{\sigma_{\text{inner1}}}{2\epsilon_0} + \frac{|\sigma_{\text{inner2}}|}{2\epsilon_0} = \frac{3\sigma/2}{2\epsilon_0} + \frac{3\sigma/2}{2\epsilon_0} = \frac{3\sigma}{2\epsilon_0}$$ Thus, the electrostatic force on +q$+q$ is: F = q E = frac3sigma q2epsilon_0$$F = q E = \frac{3\sigma q}{2\epsilon_0}$$ ### Step 1: Final Conclusion The force experienced by the point charge +q$+q$ is frac3sigma q2epsilon_0$\frac{3\sigma q}{2\epsilon_0}$. ### Pattern Recognition For conducting plates with total charges Q_1$Q_1$ and Q_2$Q_2$, always calculate the outer charge first: Q_textouter = fracQ_1+Q_22$Q_{\text{outer}} = \frac{Q_1+Q_2}{2}$. The field inside the gap is exclusively due to the inner surfaces: E = fracsigma_textinnerepsilon_0$E = \frac{\sigma_{\text{inner}}}{\epsilon_0}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula E = frack Qr^2$$E = \frac{k Q}{r^2}$$ x = fracd1 + sqrtfracq_2q_1$$x = \frac{d}{1 + \sqrt{\frac{q_2}{q_1}}}$$ ### Core Logic Let the null point where the electric field is zero be at (x, 0, 0)$(x, 0, 0)$ where 0 < x < d$0 < x < d$. At this point, the fields due to both charges are equal in magnitude and opposite in direction: frack qx^2 = frack (9q)(d - x)^2$$\frac{k q}{x^2} = \frac{k (9q)}{(d - x)^2}$$ Taking the square root on both sides: frac1x = frac3d - x implies d - x = 3x$$\frac{1}{x} = \frac{3}{d - x} \implies d - x = 3x$$ 4x = d implies x = fracd4$$4x = d \implies x = \frac{d}{4}$$ Thus, the coordinates of the null point are left(fracd4, 0, 0right)$\left(\frac{d}{4}, 0, 0\right)$. ### Step 1: Final Conclusion The point where the electric field vanishes is left(fracd4, 0, 0right)$\left(\frac{d}{4}, 0, 0\right)$. ### Pattern Recognition For two like charges, the zero-field null point always lies along the line joining them and is closer to the smaller charge. Use the standard shortcut: x = fracd1 + sqrtq_2/q_1$x = \frac{d}{1 + \sqrt{q_2/q_1}}$ measured from charge q_1$q_1$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics