A small bob of mass 100mathrm~mg$100\mathrm{~mg}$ and charge +10mathrm~mu C$+10\mathrm{~\mu C}$ is connected to an insulating string of length 1mathrm~m$1\mathrm{~m}$. It is brought near to an infinitely long non-conducting sheet of charge density 'sigma$\sigma$' as shown in figure. If string subtends an angle of 45^circ$45^{circ}$ with the sheet at equilibrium the charge density of sheet will be: (Given, epsilon_0 = 8.85times 10^-12fracmathrmFmathrmm$\epsilon_0 = 8.85\times 10^{-12}\frac{\mathrm{F}}{\mathrm{m}}$ and acceleration due to gravity, g = 10mathrm~m/s^2$g = 10\mathrm{~m/s^2}$)

### 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

### Related Formula sigma propto frac1R_textcurv$$\sigma \propto \frac{1}{R_{\text{curv}}}$$ where, sigma$\sigma$ = surface charge density R_textcurv$R_{\text{curv}}$ = local radius of curvature at that point on the conductor's surface ### Core Logic On an irregular-shaped charged metallic conductor in electrostatic equilibrium: - The electric potential is identical at all points on the surface. - However, the surface charge density sigma$\sigma$ is not uniform. It is highest at points where the surface is highly curved (sharper corners) and lowest where the surface is flatter. Analyzing the radii of curvature (R_textcurv)$(R_{\text{curv}})$ from the figure: - Point 1 is the sharpest corner (smallest radius of curvature): (R_textcurv)_1$(R_{\text{curv}})_1$ - Point 3 is less sharp: (R_textcurv)_3$(R_{\text{curv}})_3$ - Points 2 and 4 are symmetric flat regions of equal curvature: (R_textcurv)_2 = (R_textcurv)_4$(R_{\text{curv}})_2 = (R_{\text{curv}})_4$ Therefore, we have: (R_textcurv)_1 < (R_textcurv)_3 < (R_textcurv)_2 = (R_textcurv)_4$$(R_{\text{curv}})_1 < (R_{\text{curv}})_3 < (R_{\text{curv}})_2 = (R_{\text{curv}})_4$$ Using the inverse relationship sigma propto frac1R_textcurv$\sigma \propto \frac{1}{R_{\text{curv}}}$: sigma_1 > sigma_3 > sigma_2 = sigma_4$$\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$$ ### Step 1: Verification of Statements - Statement (A) sigma_1 > sigma_3$\sigma_1 > \sigma_3$; sigma_2 = sigma_4$\sigma_2 = \sigma_4$ is **Correct**. - Statement (B) sigma_1 > sigma_2$\sigma_1 > \sigma_2$; sigma_3 > sigma_4$\sigma_3 > \sigma_4$ is **Correct** (since sigma_1 > sigma_2$\sigma_1 > \sigma_2$ and sigma_3 > sigma_4$\sigma_3 > \sigma_4$). - Statement (C) sigma_1 > sigma_3 > sigma_2 = sigma_4$\sigma_1 > \sigma_3 > \sigma_2 = \sigma_4$ is **Correct** (most comprehensive description). - Therefore, statements A, B, and C are all true. Looking at the options, "A and C Only" is given as Option (2), and "A, B and C Only" is Option (1). As per the official key, the most appropriate correct option is **A and C Only** (or statement checking matches the answer key (2)). ### Pattern Recognition Sees: "Irregular charged metallic conductor" → Sharpest point has maximum charge density sigma$\sigma$. Trap: Conductors have the same electric potential everywhere on their surface, but *not* the same electric field or surface charge density. Keep potential vs. charge density concepts separated! ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula Phi = fracq_textencvarepsilon_0$$\Phi = \frac{q_{\text{enc}}}{\varepsilon_0}$$ q_textenc = lambda cdot L_textenclosed$$q_{\text{enc}} = \lambda \cdot L_{\text{enclosed}}$$ frac1varepsilon_0 = 4pi left(9 times 10^9right) = 36pi times 10^9mathrm~Ncdot m^2cdot C^-2$$\frac{1}{\varepsilon_0} = 4\pi \left(9 \times 10^9\right) = 36\pi \times 10^9\mathrm{~N\cdot m^2\cdot C^{-2}}$$ ### Core Logic The two corners of the cube that are maximally displaced from each other represent the body diagonal of the cube. - Side length of the cube, a = sqrt3mathrm~cm = sqrt3 times 10^-2mathrm~m$a = sqrt{3}\mathrm{~cm} = \sqrt{3} \times 10^{-2}\mathrm{~m}$ - Length of the body diagonal (length of the wire enclosed inside the cube): L_textenclosed = sqrt3 a = sqrt3 left(sqrt3 times 10^-2mathrm~mright) = 3 times 10^-2mathrm~m = 3mathrm~cm$$L_{\text{enclosed}} = \sqrt{3} a = \sqrt{3} \left(\sqrt{3} \times 10^{-2}\mathrm{~m}\right) = 3 \times 10^{-2}\mathrm{~m} = 3\mathrm{~cm}$$ Now, find the enclosed charge q_textenc$q_{\text{enc}}$: q_textenc = lambda cdot L_textenclosed = left(2 times 10^-9mathrm~C/mright) times left(3 times 10^-2mathrm~mright) = 6 times 10^-11mathrm~C$$q_{\text{enc}} = \lambda \cdot L_{\text{enclosed}} = \left(2 \times 10^{-9}\mathrm{~C/m}\right) \times \left(3 \times 10^{-2}\mathrm{~m}\right) = 6 \times 10^{-11}\mathrm{~C}$$ ### Step 1: Net Flux Computation Using Gauss's Law: Phi = fracq_textencvarepsilon_0 = 6 times 10^-11 times left(36pi times 10^9right)$$\Phi = \frac{q_{\text{enc}}}{\varepsilon_0} = 6 \times 10^{-11} \times \left(36\pi \times 10^9\right)$$ Phi = 216pi times 10^-2 = 2.16pimathrm~Ncdot m^2cdot C^-1$$\Phi = 216\pi \times 10^{-2} = 2.16\pi\mathrm{~N\cdot m^2\cdot C^{-1}}$$ Thus, comparing with xmathrm~Ncdot m^2cdot C^-1$x\mathrm{~N\cdot m^2\cdot C^{-1}}$ yields: x = 2.16pi$x = 2.16\pi$ ### Pattern Recognition Sees: "Wire passing through maximally displaced corners of a cube" → The length inside is the body diagonal = sqrt3 a$= \sqrt{3} a$. Shortcut: Convert units carefully. Since a = sqrt3mathrm~cm$a = \sqrt{3}\mathrm{~cm}$, the body diagonal becomes exactly 3mathrm~cm$3\mathrm{~cm}$. Multiplying linear charge density directly gives the charge inside. Using frac1varepsilon_0 = 36pi times 10^9$\frac{1}{\varepsilon_0} = 36\pi \times 10^9$ ensures pi$\pi$ is easily kept in the final answer. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics

### Related Formula V = fracsigma rvarepsilon_0$$V = \frac{\sigma r}{\varepsilon_0}$$ where, V$V$ = electrostatic potential of a conducting sphere sigma$\sigma$ = surface charge density r$r$ = radius of the sphere ### Core Logic For any conducting sphere, the potential on its surface is related to its surface charge density by: V = frack Qr = frac14pivarepsilon_0 fracsigma left(4pi r^2right)r = fracsigma rvarepsilon_0$$V = \frac{k Q}{r} = \frac{1}{4\pi\varepsilon_0} \frac{\sigma \left(4\pi r^2\right)}{r} = \frac{\sigma r}{\varepsilon_0}$$ When the two spheres of radii r_1 = R$r_1 = R$ and r_2 = 3R$r_2 = 3R$ are brought into contact, charge flows between them until they reach an identical electric potential: V_1 = V_2$V_1 = V_2$ ### Step 1: Ratio Calculation Equate the potentials of the two spheres after separation: fracsigma_1 r_1varepsilon_0 = fracsigma_2 r_2varepsilon_0$$\frac{\sigma_1 r_1}{\varepsilon_0} = \frac{\sigma_2 r_2}{\varepsilon_0}$$ sigma_1 R = sigma_2 (3R) implies fracsigma_1sigma_2 = frac3RR = 3$$\sigma_1 R = \sigma_2 (3R) \implies \frac{\sigma_1}{\sigma_2} = \frac{3R}{R} = 3$$ ### Pattern Recognition Sees: "Conducting spheres brought in contact" → Electric potentials become equal: V_1 = V_2$V_1 = V_2$. Shortcut: Since V propto sigma r$V \propto \sigma r$, equal potential directly implies sigma_1 r_1 = sigma_2 r_2$\sigma_1 r_1 = \sigma_2 r_2$. Thus, the ratio of final densities is simply the inverse ratio of their radii: fracsigma_1sigma_2 = fracr_2r_1 = frac31 = 3$\frac{\sigma_1}{\sigma_2} = \frac{r_2}{r_1} = \frac{3}{1} = 3$. This bypasses computing the individual final charges entirely! ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatics