Two point charges of 1 nC and 2 nC are placed at the two corners of equilateral triangle of side 3 cm. The work done in bringing a charge of 3 nC from infinity to the third corner of the triangle is ____ mumathrmJ$\mu\mathrm{J}$. frac14piepsilon_0=9times10^9mathrmNcdot m^2/C^2$\frac{1}{4\pi\epsilon_{0}}=9\times10^{9}\mathrm{N\cdot m^{2}/C^{2}}$

### Related Formula V = frackqr$$V = \frac{kq}{r}$$ textVolume Conservation: N cdot left(frac43pi r^3right) = frac43pi R^3$$\text{Volume Conservation: } N \cdot \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3$$ ### Core Logic From volume conservation of 3 coalescing droplets: 3 left(frac43pi r^3right) = frac43pi R^3 implies R = 3^1/3r$$3 \left(\frac{4}{3}\pi r^3\right) = \frac{4}{3}\pi R^3 \implies R = 3^{1/3}r$$ Total charge on resultant bigger bubble Q = 3q$Q = 3q$. Calculating initial potential V_i$V_i$ and final potential V_f$V_f$: V_i = frackqr$$V_i = \frac{kq}{r}$$ V_f = frack(3q)R = frac3kq3^1/3r = 3^2/3 frackqr$$V_f = \frac{k(3q)}{R} = \frac{3kq}{3^{1/3}r} = 3^{2/3} \frac{kq}{r}$$ Ratio of initial to final potential: fracV_iV_f = frac13^2/3 = 1 : 3^2/3$$\frac{V_i}{V_f} = \frac{1}{3^{2/3}} = 1 : 3^{2/3}$$

### Step 1: Final Conclusion The ratio of potentials is 1 : 3^2/3$1 : 3^{2/3}$. ### Pattern Recognition Coalescing droplets rule: For N$N$ identical drops, R = N^1/3r$R = N^{1/3}r$ and Q = Nq$Q = Nq$. Potential ratio V_i / V_f = 1 / N^2/3$V_i / V_f = 1 / N^{2/3}$. For N=3$N=3$, ratio is 1 / 3^2/3$1 / 3^{2/3}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance

### Related Formula V_textcommon = fracC_1 V_1 + C_2 V_2C_1 + C_2$$V_{\text{common}} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$$ Q_2 = C_2 V_textcommon$$Q_2 = C_2 V_{\text{common}}$$ ### Core Logic Given C_1 = 10 times 10^-6 mathrm~F, V_1 = 6.0 mathrm~V$C_1 = 10 \times 10^{-6} \mathrm{~F}, V_1 = 6.0 \mathrm{~V}$ and C_2 = 20 times 10^-6 mathrm~F, V_2 = 0 mathrm~V$C_2 = 20 \times 10^{-6} \mathrm{~F}, V_2 = 0 \mathrm{~V}$: V_textcommon = frac10^-5 times 6 + 010^-5 + 2 times 10^-5 = frac6 times 10^-53 times 10^-5 = 2 mathrm~V$$V_{\text{common}} = \frac{10^{-5} \times 6 + 0}{10^{-5} + 2 \times 10^{-5}} = \frac{6 \times 10^{-5}}{3 \times 10^{-5}} = 2 \mathrm{~V}$$ Calculating final charge on capacitor Q (C_2$C_2$): Q_2 = C_2 V_textcommon = (20 times 10^-6 mathrm~F) times 2 mathrm~V = 40 times 10^-6 mathrm~C = 4 times 10^-5 mathrm~C$$Q_2 = C_2 V_{\text{common}} = (20 \times 10^{-6} \mathrm{~F}) \times 2 \mathrm{~V} = 40 \times 10^{-6} \mathrm{~C} = 4 \times 10^{-5} \mathrm{~C}$$ Comparing with alpha times 10^-5 mathrm~C implies alpha = 4$\alpha \times 10^{-5} \mathrm{~C} \implies \alpha = 4$. ### Step 1: Final Conclusion The value of alpha$\alpha$ is 4. ### Pattern Recognition Charge distribution rule: Total initial charge Q_total = C_1 V_1 = 60\,mumathrmC$Q_{total} = C_1 V_1 = 60\,\mu\mathrm{C}$. Final charge splits in proportion to capacitance ratio C_2 / (C_1+C_2) = 2/3$C_2 / (C_1+C_2) = 2/3$. Q_2 = (2/3) times 60\,mumathrmC = 40\,mumathrmC = 4 times 10^-5mathrm~C$Q_2 = (2/3) \times 60\,\mu\mathrm{C} = 40\,\mu\mathrm{C} = 4 \times 10^{-5}\mathrm{~C}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance

### Related Formula 1. Uniform Electric Field between parallel plates: E = fracVd$$E = \frac{V}{d}$$ 2. Potential difference between two points along the direction of the field: Delta V = E cdot Delta x$$\Delta V = E \cdot \Delta x$$ ### Core Logic The separation between the plates is d = 10 \ mathrmcm$d = 10 \ \mathrm{cm}$. The electric field vecE$\vec{E}$ between them is uniform and runs perpendicularly between the plates (along the horizontal direction, +x$+x$): E = fracV10 \ mathrmV/cm$$E = \frac{V}{10} \ \mathrm{V/cm}$$ From the geometry shown in the figure: - AC = 3 \ mathrmcm$AC = 3 \ \mathrm{cm}$ is perpendicular to vecE$\vec{E}$ (vertical direction). - CB = 4 \ mathrmcm$CB = 4 \ \mathrm{cm}$ is parallel to vecE$\vec{E}$ (horizontal direction). ### Step 1: Calculate potential difference between A and B Since line AC$AC$ is perpendicular to vecE$\vec{E}$: V_A - V_C = 0 implies V_A = V_C$$V_A - V_C = 0 \implies V_A = V_C$$ Thus, the potential difference between A$A$ and B$B$ is completely due to the horizontal displacement CB$CB$: V_AB = V_A - V_B = E cdot (CB)$$V_{AB} = V_A - V_B = E \cdot (CB)$$ Substitute the values: V_AB = left(fracV10right) times 4 = frac25 V$$V_{AB} = \left(\frac{V}{10}\right) \times 4 = \frac{2}{5} V$$ Thus, the potential difference between A$A$ and B$B$ is frac25V$\frac{2}{5}V$. ### Pattern Recognition Sees: Uniform electric field, potential difference over diagonal paths. Trap: Projecting along the hypotenuse length (5text cm$5\text{ cm}$) directly without considering the electric field's physical direction. Shortcut: Electric field is purely horizontal. Hence, only horizontal displacement matters. The horizontal displacement is 4text cm$4\text{ cm}$ out of 10text cm$10\text{ cm}$ total plate gap, so the potential difference is frac410V = frac25V$\frac{4}{10}V = \frac{2}{5}V$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance

### Related Formula When a charged capacitor C_1$C_1$ with voltage V_1$V_1$ is connected in parallel to an uncharged capacitor C_2$C_2$, the common final potential V_c$V_c$ is determined by the conservation of charge: V_c = fracC_1 V_1 + C_2 V_2C_1 + C_2$$V_c = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$$ Since C_2$C_2$ is initially uncharged (V_2 = 0$V_2 = 0$): V_c = fracC_1 V_1C_1 + C_2$$V_c = \frac{C_1 V_1}{C_1 + C_2}$$ ### Core Logic Given parameters: - First capacitor C_1 = 100mathrm~pF$C_1 = 100\mathrm{~pF}$ - Initial potential V_1 = 60mathrm~V$V_1 = 60\mathrm{~V}$ - Common final voltage V_c = 20mathrm~V$V_c = 20\mathrm{~V}$ ### Step 1: Solve for C_2$C_2$ Substitute the values into the common potential expression: 20 = frac100 times 60100 + C_2$$20 = \frac{100 \times 60}{100 + C_2}$$ 20(100 + C_2) = 6000$$20(100 + C_2) = 6000$$ 100 + C_2 = frac600020 = 300$$100 + C_2 = \frac{6000}{20} = 300$$ C_2 = 300 - 100 = 200mathrm~pF$$C_2 = 300 - 100 = 200\mathrm{~pF}$$ ### Pattern Recognition Think of common potential as dilution. The potential drops to 1/3$1/3$ of its initial value (20mathrm~V/60mathrm~V$20\mathrm{~V}/60\mathrm{~V}$). This requires the total capacitance to triple (3 times C_1$3 \times C_1$). Since they are in parallel, the added capacitor must be 2 times C_1 = 200mathrm~pF$2 \times C_1 = 200\mathrm{~pF}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance