### 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

### Related Formula The change in potential energy Delta U$\Delta U$ of a charge q$q$ moved between two points of potentials V_C$V_C$ and V_D$V_D$ is: Delta U = q_3 (V_D - V_C)$\Delta U = q_3 (V_D - V_C)$ Potential due to a point charge q$q$ at distance r$r$ is: V = frac14piepsilon_0 fracqr$V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$ ### Core Logic Let point A$A$ hold charge q_1$q_1$ and B$B$ hold q_2$q_2$ separated by 30 mathrm~cm = 0.3 mathrm~m$30 \mathrm{~cm} = 0.3 \mathrm{~m}$. - The path of q_3$q_3$ is a circular arc of radius R = 40 mathrm~cm = 0.4 mathrm~m$R = 40 \mathrm{~cm} = 0.4 \mathrm{~m}$ centered at A$A$. Thus, distance of C$C$ and D$D$ from q_1$q_1$ is constant: r_1C = r_1D = 0.4 mathrm~m$r_{1C} = r_{1D} = 0.4 \mathrm{~m}$ - Distance of C$C$ from q_2$q_2$ (B$B$): r_2C = sqrtAC^2 + AB^2 = sqrt40^2 + 30^2 = 50 mathrm~cm = 0.5 mathrm~m$r_{2C} = \sqrt{AC^2 + AB^2} = \sqrt{40^2 + 30^2} = 50 \mathrm{~cm} = 0.5 \mathrm{~m}$ - Distance of D$D$ from q_2$q_2$ (B$B$): r_2D = AD - AB = 40 - 30 = 10 mathrm~cm = 0.1 mathrm~m$r_{2D} = AD - AB = 40 - 30 = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$ ### Step 1: Calculate Potentials Potential at $ ### Step 1: Calculate Potentials Potential at $C due to $ due to $q_1 and $ and $q_2: $: $V_C = frac14piepsilon_0 left( fracq_10.4 + fracq_20.5 right)$V_C = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.5} \right)$ Potential at $ Potential at $D due to $ due to $q_1 and $ and $q_2: $: $V_D = frac14piepsilon_0 left( fracq_10.4 + fracq_20.1 right)$V_D = \frac{1}{4\pi\epsilon_0} \left( \frac{q_1}{0.4} + \frac{q_2}{0.1} \right)$ ### Step 2: Difference in Potential Energy The potential difference is: $ ### Step 2: Difference in Potential Energy The potential difference is: $V_D - V_C = frac14piepsilon_0 left( fracq_20.1 - fracq_20.5 right) = fracq_24piepsilon_0 [10 - 2] = frac8q_24piepsilon_0$V_D - V_C = \frac{1}{4\pi\epsilon_0} \left( \frac{q_2}{0.1} - \frac{q_2}{0.5} \right) = \frac{q_2}{4\pi\epsilon_0} [10 - 2] = \frac{8q_2}{4\pi\epsilon_0}$ (Notice that the potential contribution of $ (Notice that the potential contribution of $q_1 cancels out because $ cancels out because $C and $ and $D are equidistant from $ are equidistant from $q_1). The change in potential energy is: $). The change in potential energy is: $Delta U = q_3 (V_D - V_C) = fracq_3 (8q_2)4piepsilon_0$\Delta U = q_3 (V_D - V_C) = \frac{q_3 (8q_2)}{4\pi\epsilon_0}$ Comparing with $ Comparing with $\frac{q_3 K}{4\pi\epsilon_0} yields $ yields $K = 8q_2. ### Pattern Recognition Sees: Arc path centered on one of the charges. Shortcut: Since the path is circular about $. ### Pattern Recognition Sees: Arc path centered on one of the charges. Shortcut: Since the path is circular about $q_1, $, $q_1 contributes nothing to the potential difference between the end points. The entire change in potential energy is due to $ contributes nothing to the potential difference between the end points. The entire change in potential energy is due to $q_2. The distance difference translates to potential difference $. The distance difference translates to potential difference $\Delta V = q_2 \left(\frac{1}{0.1} - \frac{1}{0.5}\right) = 8q_2, so $, so $K = 8q_2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance

### Related Formula Q_texttotal = C_1 V_0$Q_{\text{total}} = C_1 V_0$ V_c = fracQ_texttotalC_1 + C_2$V_c = \frac{Q_{\text{total}}}{C_1 + C_2}$ q = C cdot V_c$q = C \cdot V_c$ ### Core Logic 1. **Initial Charge Calculation**: Before closing the switch, capacitor C_1$C_1$ accumulates total charge: q_1' = C_1 cdot V_0 = 6mumathrmF times 5mathrmV = 30mumathrmC$q_1' = C_1 \cdot V_0 = 6\mu\mathrm{F} \times 5\mathrm{V} = 30\mu\mathrm{C}$ Uncharged capacitor C_2$C_2$ holds q_2' = 0$q_2' = 0$.

2. **Redistribution at Equilibrium**: Closing the switch causes parallel redistribution until they reach a common potential V_c$V_c$:

V_c = frac30mumathrmC + 06mumathrmF + 12mumathrmF = frac3018 = frac53mathrm~V$V_c = \frac{30\mu\mathrm{C} + 0}{6\mu\mathrm{F} + 12\mu\mathrm{F}} = \frac{30}{18} = \frac{5}{3}\mathrm{~V}$ 3. **Final Charges**: q_1 = C_1 cdot V_c = 6 times frac53 = 10mumathrmC$q_1 = C_1 \cdot V_c = 6 \times \frac{5}{3} = 10\mu\mathrm{C}$ q_2 = C_2 cdot V_c = 12 times frac53 = 20mumathrmC$q_2 = C_2 \cdot V_c = 12 \times \frac{5}{3} = 20\mu\mathrm{C}$ Hence, the charges are 10mumathrmC$10\mu\mathrm{C}$ and 20mumathrmC$20\mu\mathrm{C}$ respectively. ### Pattern Recognition For parallel combination setups, the final charge splits in the exact direct ratio of their capacitances: q_1 : q_2 = C_1 : C_2 = 6 : 12 = 1 : 2$q_1 : q_2 = C_1 : C_2 = 6 : 12 = 1 : 2$. Out of 30mumathrmC$30\mu\mathrm{C}$, the breakdown must cleanly be 10mumathrmC$10\mu\mathrm{C}$ and 20mumathrmC$20\mu\mathrm{C}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance

### Related Formula mathrmV = textsame$\mathrm{V} = \text{same}$ mathrmU = frac12 mathrmCmathrmV^2$\mathrm{U} = \frac{1}{2} \mathrm{C}\mathrm{V}^2$ mathrmq = mathrmCmathrmV$\mathrm{q} = \mathrm{C}\mathrm{V}$ ### Core Logic Since both capacitors are connected in parallel across the same battery, their potential difference mathrmV$\mathrm{V}$ is identical. From the given charge-time graph, at any specific instant of time, the charge accumulated on the second capacitor is greater than that on the first: mathrmq_2 > mathrmq_1$\mathrm{q}_2 > \mathrm{q}_1$ Using the relation mathrmq = mathrmCmathrmV$\mathrm{q} = \mathrm{C}\mathrm{V}$, since mathrmV$\mathrm{V}$ is identical, we get: mathrmC_2 > mathrmC_1$\mathrm{C}_2 > \mathrm{C}_1$ Now, the electrostatic energy stored in a capacitor is given by: mathrmU = frac12 mathrmCmathrmV^2$\mathrm{U} = \frac{1}{2} \mathrm{C}\mathrm{V}^2$ Since mathrmC_2 > mathrmC_1$\mathrm{C}_2 > \mathrm{C}_1$ and mathrmV$\mathrm{V}$ is constant, the stored energy satisfies: mathrmU_2 > mathrmU_1$\mathrm{U}_2 > \mathrm{U}_1$ ### Step 1: Final Conclusion Thus, the correct relationships are mathrmC_2 > mathrmC_1$\mathrm{C}_2 > \mathrm{C}_1$ and mathrmU_2 > mathrmU_1$\mathrm{U}_2 > \mathrm{U}_1$. ### Pattern Recognition Sees parallel connection → Immediately lock potential difference mathrmV$\mathrm{V}$ as constant. This simplifies mathrmq propto mathrmC$\mathrm{q} \propto \mathrm{C}$ and mathrmU propto mathrmC$\mathrm{U} \propto \mathrm{C}$, creating a direct linear bridge from graph height to energy capacity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electrostatic Potential and Capacitance