Two large plane parallel conducting plates are kept 10mathrmcm$10\mathrm{cm}$ apart as shown in figure. The potential difference between them is V. The potential difference between the points A and B (shown in the figure) is :
The diagram shows two parallel plates separated by 10 cm, with points A and B defining a right triangle of legs 3 cm and 4 cm.
A.frac14 mathrm~V$\frac{1}{4} \mathrm{~V}$
B.frac25 mathrm~V$\frac{2}{5} \mathrm{~V}$
C.frac34 mathrm~V$\frac{3}{4} \mathrm{~V}$
D.1 mathrm~V$1 \mathrm{~V}$
Solution & Explanation
### 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
The diagram shows two parallel plates separated by 10 cm, with points A and B defining a right triangle of legs 3 cm and 4 cm.
Keywords:#potential difference between points parallel plates#JEE Main 2025 Evening Q9#electrostatic potential uniform electric field#conducting plates electric potential gradient#parallel plates#conducting plates#electric potential gradient
More Electrostatic Potential and Capacitance Previous-Year Questions
Q30jee_main_2026_22_january_eveningElectric Potential of Coalescing Bubbles
Three small identical bubbles of water having same charge on each coalesce to form a bigger bubble. Then the ratio of the potentials on one initial bubble and that on the resultant bigger bubble is :
A.1:3^1/3$1:3^{1/3}$
B.1:2^2/3$1:2^{2/3}$
C.3^2/3:1$3^{2/3}:1$
D.1:3^2/3$1:3^{2/3}$
Solution
### 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$$\frac{V_i}{V_f} = \frac{1}{3^{2/3}} = 1 : 3^{2/3}$$Coalescing charged bubbles diagram for Q30 - JEE Main 2026 Evening
### 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
Q48jee_main_2026_22_january_eveningSharing of Charges between Capacitors
A capacitor P with capacitance 10 times 10^-6$10 \times 10^{-6}$ F is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor P is connected across another capacitor Q with capacitance 20 times 10^-6$20 \times 10^{-6}$ F. The charge on capacitor Q when equilibrium is established will be alpha times 10^-5$\alpha \times 10^{-5}$ C (assume capacitor Q does not have any charge initially), the value of alpha$\alpha$ is ____.
Numerical Answer.Answer: 4 to 4
Solution
### 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
Q44jee_main_2026_28_january_morningPotential Energy of System of Charges
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$\frac{1}{4\pi\epsilon_{0}}=9\times10^{9}\mathrm{N\cdot m^{2}/C^{2}}$
A.2.7$2.7$
B.5.4$5.4$
C.3.3$3.3$
D.27$27$
Solution
### Related Formula
W = q_3 cdot V_textsystem$$W = q_3 \cdot V_{\text{system}}$$V = frackq_1r_1 + frackq_2r_2$$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$$
### Core Logic
Work done is equal to the potential energy acquired by the 3 nC charge in the electric field of the other two charges.
Equilateral triangle charge diagram
### Step 1: Formula Setup
W = left( frackq_1ell + frackq_2ell right) q_3$$W = \left( \frac{kq_1}{\ell} + \frac{kq_2}{\ell} \right) q_3$$
### Step 2: Value Substitution
q_1 = 1 times 10^-9 mathrm~C$$q_1 = 1 \times 10^{-9} \mathrm{~C}$$q_2 = 2 times 10^-9 mathrm~C$$q_2 = 2 \times 10^{-9} \mathrm{~C}$$q_3 = 3 times 10^-9 mathrm~C$$q_3 = 3 \times 10^{-9} \mathrm{~C}$$ell = 3 times 10^-2 mathrm~m$$\ell = 3 \times 10^{-2} \mathrm{~m}$$W = frac9 times 10^93 times 10^-2 left( 1 times 10^-9 + 2 times 10^-9 right) times 3 times 10^-9$$W = \frac{9 \times 10^9}{3 \times 10^{-2}} \left( 1 \times 10^{-9} + 2 \times 10^{-9} \right) \times 3 \times 10^{-9}$$
### Step 3: Final Calculation
W = 3 times 10^11 times 3 times 10^-9 times 3 times 10^-9$$W = 3 \times 10^{11} \times 3 \times 10^{-9} \times 3 \times 10^{-9}$$W = 27 times 10^-7 mathrm~J = 2.7 times 10^-6 mathrm~J = 2.7 \,mumathrmJ$$W = 27 \times 10^{-7} \mathrm{~J} = 2.7 \times 10^{-6} \mathrm{~J} = 2.7 \,\mu\mathrm{J}$$
### Pattern Recognition
Factor out common terms (k/ell$k/\ell$) before substituting micro/nano orders to prevent simple arithmetic exponent errors.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Physics: Electrostatic Potential and Capacitance
Q8jee_main_2025_03_april_eveningCombination of Capacitors and Charge Sharing
Using a battery, a 100mathrm~pF$100\mathrm{~pF}$ capacitor is charged to 60mathrm~V$60\mathrm{~V}$ and then the battery is removed. After that, a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is 20mathrm~V$20\mathrm{~V}$, its capacitance is: (in pF)
A. 600
B. 200
C. 400
D. 100
Solution
### 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
More Electrostatic Potential and Capacitance Questions — jee_main_2025_02_april_evening
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