### 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

### Related Formula E = fracsigma2epsilon_0 quad text(field of infinite non-conducting charged sheet)$E = \frac{\sigma}{2\epsilon_0} \quad \text{(field of infinite non-conducting charged sheet)}$ tantheta = fracF_emg$\tan\theta = \frac{F_e}{mg}$ ### Core Logic In equilibrium, three forces act on the suspended charged bob: 1. Tension T$T$ directed along the string at theta = 45^circ$\theta = 45^{circ}$ with the vertical sheet. 2. Weight mg$mg$ directed vertically downwards. 3. Electrostatic repulsion force F_e = qE$F_e = qE$ acting horizontally away from the sheet. From the balance of forces in vertical and horizontal directions: T cos(45^circ) = mg$T \cos(45^{circ}) = mg$ T sin(45^circ) = q E$T \sin(45^{circ}) = q E$ Dividing the two equations: tan(45^circ) = fracq Emg = 1 implies q E = mg$\tan(45^{circ}) = \frac{q E}{mg} = 1 \implies q E = mg$ Substitute the expression for E$E$: q left(fracsigma2epsilon_0right) = mg implies sigma = frac2 epsilon_0 m gq$q \left(\frac{\sigma}{2\epsilon_0}\right) = mg \implies \sigma = \frac{2 \epsilon_0 m g}{q}$ Now plug in the given numerical values: - m = 100mathrm~mg = 100 times 10^-6mathrm~kg = 10^-4mathrm~kg$m = 100\mathrm{~mg} = 100 \times 10^{-6}\mathrm{~kg} = 10^{-4}\mathrm{~kg}$ - q = +10mathrm~mu C = 10 times 10^-6mathrm~C = 10^-5mathrm~C$q = +10\mathrm{~\mu C} = 10 \times 10^{-6}\mathrm{~C} = 10^{-5}\mathrm{~C}$ - g = 10mathrm~m/s^2$g = 10\mathrm{~m/s^2}$ - epsilon_0 = 8.85 times 10^-12mathrm~F/m$\epsilon_0 = 8.85 \times 10^{-12}\mathrm{~F/m}$ sigma = frac2 times (8.85 times 10^-12) times 10^-4 times 1010^-5$\sigma = \frac{2 \times (8.85 \times 10^{-12}) \times 10^{-4} \times 10}{10^{-5}}$ sigma = 17.7 times 10^-10mathrm~C/m^2 = 1.77 times 10^-9mathrm~C/m^2 = 1.77mathrm~nC/m^2$\sigma = 17.7 \times 10^{-10}\mathrm{~C/m}^2 = 1.77 \times 10^{-9}\mathrm{~C/m}^2 = 1.77\mathrm{~nC/m}^2$ ### Step 1: Final Conclusion The charge density of the sheet is 1.77mathrm~nC/m^2$1.77\mathrm{~nC/m}^2$. ### Pattern Recognition For a charge hanging near a vertical charged sheet, the equilibrium angle is governed by tantheta = fracF_emg$\tan\theta = \frac{F_e}{mg}$. For theta = 45^circ$\theta = 45^{circ}$, the horizontal force equals the vertical force (F_e = mg$F_e = mg$). Be careful to use the field of a non-conducting sheet: E = fracsigma2epsilon_0$E = \frac{\sigma}{2\epsilon_0}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula W_A rightarrow B = q(V_B - V_A)$W_{A \rightarrow B} = q(V_B - V_A)$ where, W_A rightarrow B$W_{A \rightarrow B}$ = work done in moving a test charge q$q$ from point A$A$ to B$B$ V_A, V_B$V_A, V_B$ = electrostatic potentials at points A$A$ and B$B$ ### Core Logic For a uniformly charged spherical shell of radius R$R$ and charge Q$Q$, the electric field inside the shell is zero (E = 0$E = 0$). Consequently, the electric potential V$V$ remains constant throughout the interior of the shell and equals its value on the surface: V_textinside = V_textsurface = frac14pivarepsilon_0 fracQR$V_{\text{inside}} = V_{\text{surface}} = \frac{1}{4\pi\varepsilon_0} \frac{Q}{R}$ Since the potential is identical at all interior points (V_A = V_B$V_A = V_B$), the potential difference is zero: Delta V = V_B - V_A = 0$\Delta V = V_B - V_A = 0$ Thus, the work done in moving any test charge inside is strictly zero: W = q Delta V = 0$W = q \Delta V = 0$ ### Step 1: Analyzing the Statements 1. **Assertion A**: "Work done in moving a test charge inside is zero..." - This is **True**. 2. **Reason R**: "Electrostatic potential inside is constant and same as on the surface..." - This is **True** and directly explains why the potential difference Delta V = 0$\Delta V = 0$, making the work done zero. Therefore, both statements are true and R$R$ is the correct explanation of A$A$. ### Pattern Recognition Sees: "Uniformly charged spherical shell" + "Work done inside" → Potential difference Delta V = 0 implies W = 0$\Delta V = 0 \implies W = 0$. Shortcut: Since E_textinside = 0$E_{\text{inside}} = 0$, potential inside is flat/constant. No potential difference means zero work. Both statements are true and connected. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics