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Electric Potential & Capacitance Formulas for JEE 2027
Electric Potential and Capacitance formulas cover electric potential due to charges, potential energy, equipotential surfaces, capacitor combinations, dielectrics and energy storage. This formula sheet brings together important relations, their conditions, worked examples and revision tips to help you understand and solve questions for JEE Main and Advanced.
Use this sheet while practising JEE Capacitance Questions and JEE Electric Potential and Capacitance PYQs. Focus on charge signs, SI units and whether a capacitor remains connected to a battery, as these details determine which formula applies. Combine regular practice with JEE Main previous papers, and revisit the quick revision section to refresh key concepts before mock tests and the exam.
Notation
| Symbol | Stands for | Units |
|---|---|---|
| $$k$$ | Coulomb constant $$1/4\pi \varepsilon_0$$ | $$\text{N m}^2\text{C}^{-2}$$ |
| $$\varepsilon_0$$ | Permittivity of free space | $$\text{F m}^{-1}$$ |
| $$q,Q$$ | Point charge | $$\text{C}$$ |
| $$r$$ | Distance from charge to field point | $$\text{m}$$ |
| $$V$$ | Electric potential (absolute) | $$\text{V}$$ |
| $$\Delta V$$ | Potential difference | $$\text{V}$$ |
| $$U$$ | Electric potential energy | $$\text{J}$$ |
| $$\vec E$$ | Electric field | $$\text{N C}^{-1}$$ |
| $$\lambda, \sigma, \rho$$ | Linear, surface, volume charge density | $$\text{C m}^{-1}, \text{C m}^{-2}, \text{C m}^{-3}$$ |
| $$C$$ | Capacitance | $$\text{F}$$ |
| $$A$$ | Plate area | $$\text{m}^2$$ |
| $$d$$ | Plate separation | $$\text{m}$$ |
| $$\kappa$$ | Dielectric constant $$\varepsilon/\varepsilon_0$$ | dimensionless |
| $$u$$ | Energy density | $$\text{J m}^{-3}$$ |
| $$R$$ | Radius (sphere/cylinder) | $$\text{m}$$ |
| $$V_0$$ | Initial potential or supply voltage | $$\text{V}$$ |
| $$\tau$$ | RC time constant | $$\text{s}$$ |
Electric Potential due to Point Charges Formulas
This is the scalar cousin of Coulomb’s vector law. You trade direction for sign, sum easily, and reach for it whenever the question shows discrete charges or wants work done in moving a charge. It sits right after field calculation because potential is the line integral of field; knowing it here lets you step into continuous distributions next.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$V = k\frac{q}{r}$$ | Potential at a point due to a single charge $$q$$ | Reference at infinity, medium vacuum/air |
| $$V = \sum_i k\frac{q_i}{r_i}$$ | Net potential of a system of point charges | Superposition holds |
| $$U = k\frac{q_1 q_2}{r}$$ | Potential energy of two‐charge system | Point charges, $$r$$ centre separation |
| $$U = \frac12 \sum_i q_i V_i$$ | Total potential energy of many charges | Each $$V_i$$ excludes $$q_i$$ itself |
| $$\vec E = -\nabla V$$ | Field from potential gradient | Cartesian/any coordinates |
| $$\Delta W = -q\Delta V$$ | Work done by electrostatic force | Quasi-static move |
Worked Example
Three charges $$2\text{ µC}$$, $$-3\text{ µC}$$ and $$4\text{ µC}$$ are placed at the corners of an equilateral triangle of side $$0.2\text{ m}$$. Find the potential at the centroid.
- Distance from each corner to centroid $$r = \frac{\sqrt3}{3}a = \frac{0.2\sqrt3}{3}=0.115\text{ m}$$.
- Compute each term: $$V = k\frac{1}{0.115}\left(2\times10^{-6}-3\times10^{-6}+4\times10^{-6}\right).$$
- Net charge inside parenthesis $$= 3\times10^{-6}\text{ C}$$.
- $$k = 9\times10^{9}$$, so $$V = 9\times10^{9}\times\frac{3\times10^{-6}}{0.115}= \frac{27\times10^{3}}{0.115}\approx2.35\times10^{5}\text{ V}.$$
Answer: $$2.3\times10^{5}\text{ V}$$
Shortcuts and Special Cases
- At midpoint of two equal opposite charges (dipole), $$V=0$$ even though $$\vec E\neq0$$.
- Inside a uniformly charged thin spherical shell, $$V=$$ constant $$= kQ/R$$. No field, but potential is not zero.
- Potential of identical charges at vertices of regular polygon at centre $$= nkq/R$$ where $$n$$ is number of charges.
Watch Out
- For potential energy of multiple charges, do not sum $$kq_i/r_i$$ again – include the 1/2 factor.
- Keep micro-coulomb to coulomb: $$1\text{ µC}=10^{-6}\text{ C}$$.
- Sign of work: moving against field increases $$V$$, work done by you is $$+q\Delta V$$.
Potential of Continuous Charge Distributions Formulas
After single charges, JEE loves rods, rings and discs because integration stays single‐variable. You integrate $$dV$$ not $$d\vec E$$, so symmetry needed is milder. Questions usually compare on‐axis potential with field, or ask work to bring a test charge.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$dV = k\frac{dq}{r}$$ | Elemental potential | Reference at infinity |
| Ring on axis: $$V = k\frac{Q}{\sqrt{R^{2}+x^{2}}}$$ | Potential at distance $$x$$ from centre | Thin ring, radius $$R$$ |
| Long line: $$V = 2k\lambda \ln\frac{r_2}{r_1}$$ | Potential difference between two radial points | Infinite line charge |
| Disc on axis: $$V = 2k\sigma\left(\sqrt{R^{2}+x^{2}}-x\right)$$ | Potential at point on axis | Uniform surface charge |
| Sphere (solid): $$V = k\frac{Q}{2R}\left(3-\frac{r^{2}}{R^{2}}\right)$$ | Potential at internal point $$r\le R$$ | Uniform volume charge |
| Outside any distribution: $$V = k\frac{Q_{tot}}{r}$$ | Equivalent point charge result | Condition $$r \gt$$ greatest dimension |
Worked Example
A uniformly charged disc of radius $$0.1\text{ m}$$ carries $$5\text{ mC}$$ total charge. Find the potential at a point $$0.06\text{ m}$$ on its axis.
- Surface charge $$\sigma = Q/(\pi R^{2})=5\times10^{-3}/(\pi 0.1^{2})=0.159\text{ C m}^{-2}$$.
- Use disc formula: $$V = 2k\sigma\left(\sqrt{R^{2}+x^{2}}-x\right).$$
- Calculate inside: $$\sqrt{0.1^{2}+0.06^{2}}= \sqrt{0.01+0.0036}= \sqrt{0.0136}=0.1166\text{ m}.$$
- Bracket term $$=0.1166-0.06=0.0566\text{ m}.$$
- $$V = 2\times 9\times10^{9}\times0.159\times0.0566 \approx 1.62\times10^{8}\text{ V}.$$
Answer: $$1.6\times10^{8}\text{ V}$$
Shortcuts and Special Cases
- On axis far away $$x \gg R$$, ring behaves like point charge: $$V \approx kQ/x$$.
- Potential inside uniformly charged sphere is quadratic in $$r$$, field is linear. Remember their graphs.
- For JEE Advanced, they sometimes differentiate the disc potential to derive field – keep $$E_x = -dV/dx$$ ready.
Watch Out
- Line charge potential diverges; examiner always asks potential difference, never absolute $$V$$.
- Don’t mix up disc and ring formulas – the factor 2 and the minus $$x$$ trip many.
- Use metres for all lengths before plugging into 9 ×10⁹.
Equipotential Surfaces & Potential Gradient Formulas
Graphical part of the topic: you translate numbers into surfaces and slopes. JEE uses it for conceptual MCQs: direction of motion, work along closed path, relation between $$E$$ and spacing of surfaces. It ties previous potentials to next chapter of capacitance.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$\vec E = -\nabla V$$ | Field from potential | — |
| $$E = -\frac{dV}{dr}$$ | Magnitude in radial symmetry | Spherical cases |
| Work along equipotential $$=0$$ | Energy statement | Static field |
| Density of surfaces $$\propto E$$ | Qualitative relation | Field sketching |
Worked Example
The potential in a region is $$V(x) = 5x^{2}+3$$ (in volts, $$x$$ in metres). Find the electric field at $$x = 0.4\text{ m}$$ and work done in moving a $$2\text{ C}$$ charge from $$x=0.4\text{ m}$$ to $$x=0.2\text{ m}$$.
- Field $$E_x = -dV/dx = -10x$$.
- At $$x=0.4$$, $$E_x = -10(0.4) = -4\text{ V m}^{-1}$$ (towards negative $$x$$).
- Potential difference $$\Delta V = V(0.2)-V(0.4)=5(0.04)-5(0.16)=0.2-0.8=-0.6\text{ V}$$.
- Work done by field $$= -q\Delta V = -2(-0.6)=1.2\text{ J}$$.
Answer: $$E=-4\text{ V m}^{-1}, \; W=1.2\text{ J}$$
Shortcuts and Special Cases
- If potential is quadratic, field is linear; if potential is linear, field is constant.
- In any conductor in electrostatic equilibrium, the entire volume is a single equipotential.
Watch Out
- Minus sign: $$\vec E$$ points from higher to lower $$V$$.
- Units: field in $$\text{V m}^{-1}$$ is numerically equal to $$\text{N C}^{-1}$$.
Capacitance of an Isolated Conductor Formulas
Capacitance starts before you even bring two plates together. JEE sometimes asks the $$4\pi \varepsilon_0 R$$ value for a sphere or energy stored on an isolated sphere connected to source. This bridges potential to the device world of capacitors.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$C = \frac{Q}{V}$$ | Definition of capacitance | Linear regime |
| Isolated sphere: $$C = 4\pi \varepsilon_0 R$$ | Capacitance of single conducting sphere | Vacuum/air |
| Energy: $$U = \frac{Q^{2}}{2C} = \frac12 CV^{2}$$ | Energy stored | Static charge |
| Potential change when charge added: $$\Delta V = \Delta Q/C$$ | Small change analysis | — |
Worked Example
A conducting sphere of radius $$0.15\text{ m}$$ is charged to $$12\text{ kV}$$. Find the charge stored and energy in it.
- Capacitance $$C = 4\pi \varepsilon_0 R = 4\pi (8.85\times10^{-12})\times0.15 =1.67\times10^{-11}\text{ F}$$.
- Charge $$Q = CV = 1.67\times10^{-11}\times1.2\times10^{4}=2.00\times10^{-7}\text{ C}$$.
- Energy $$U = \frac12 CV^{2}=0.5\times1.67\times10^{-11}\times(1.2\times10^{4})^{2}=1.2\times10^{-3}\text{ J}$$.
Answer: $$Q=2.0\times10^{-7}\text{ C},\; U=1.2\text{ mJ}$$
Shortcuts and Special Cases
- Capacitance depends only on geometry and medium, not on $$Q$$ or $$V$$.
- Changing radius by 1% changes $$C$$ by 1%. Linear proportion directly helps quick ratio questions.
Watch Out
- Do not use $$C = \varepsilon_0 A/d$$ for isolated sphere – that is plate formula.
- Capacitance of earth ≈$$ 710\;\mu\text{F}$$ (radius 6400 km) – nice trivia, sometimes appears.
Standard Parallel-Plate, Cylindrical & Spherical Capacitors Formulas
Most numerical capacitors in exams come in one of these three geometries. You memorise each expression once and keep an eye on what is fixed – plate area, separation or radii. Links directly to dielectric insertion next.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Parallel plate: $$C = \frac{\varepsilon_0 A}{d}$$ | Capacitance of two large plates | Edge effects neglected, $$d \ll \sqrt{A}$$ |
| Cylindrical: $$C = \frac{2\pi \varepsilon_0 L}{\ln(b/a)}$$ | Coaxial cylinders radius $$a,b$$ length $$L$$ | $$L \gg b$$ |
| Spherical (concentric): $$C = \frac{4\pi \varepsilon_0 ab}{b-a}$$ | Radii $$a,b$$ (inner, outer) | Filled with air |
| With dielectric: replace $$\varepsilon_0 \to \kappa\varepsilon_0$$ | Capacitance increases by $$\kappa$$ | Homogeneous dielectric |
| Field between plates: $$E = \frac{V}{d}$$ | Uniform field magnitude | Neglect edge |
| Surface charge: $$\sigma = \varepsilon_0 E$$ | Charge density on each plate | No dielectric |
Worked Example
A coaxial cable has inner conductor radius $$1\text{ mm}$$, outer conductor radius $$5\text{ mm}$$ and dielectric of permittivity $$\kappa=2.5$$. Find capacitance per metre.
- Plug values: $$C/L = \frac{2\pi \kappa \varepsilon_0}{\ln(b/a)}.$$ $$\ln(b/a)=\ln(5/1)=\ln5=1.609.$$
- $$C/L=2\pi\times2.5\times8.85\times10^{-12}/1.609= 86.5\times10^{-12}\text{ F m}^{-1}=86.5\text{ pF m}^{-1}.$$
Answer: $$87\text{ pF m}^{-1}$$
Shortcuts and Special Cases
- Thin spherical capacitor $$b-a\ll a$$ behaves like plate: $$C\approx4\pi\varepsilon_0 a^{2}/(b-a).$$
- Parallel plate doubling area doubles $$C$$; halving distance doubles $$C$$ – a common quick ratio asked.
Watch Out
- Use natural log in coax formula, not log base 10.
- Units: convert mm to m before substitution.
- In spherical capacitor, $$a$$ must be inner radius; using outer swaps sign in $$\ln$$.
Combination of Capacitors Formulas
Once capacitances are known, networks mirror resistor algebra but with role of series and parallel swapped for charges and voltages. JEE slips mixed combinations inside circuits; you must read what is common – potential or charge.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Series: $$\frac1{C_s} = \sum \frac1{C_i}$$ | Equivalent capacitance | Same charge on each |
| Parallel: $$C_p = \sum C_i$$ | Equivalent capacitance | Same potential across each |
| Charge division in parallel: $$Q_i = C_i V$$ | Individual charges | Common $$V$$ |
| Voltage division in series: $$V_i = Q/C_i$$ | Voltage across each | Common $$Q$$ |
| Energy additivity: $$U_{tot}=\sum U_i$$ | Total stored energy | No mutual interaction |
Worked Example
Three capacitors $$2\text{ µF}, 3\text{ µF}, 6\text{ µF}$$ are connected in series to a $$120\text{ V}$$ source. Find (a) equivalent capacitance, (b) charge on each, (c) potential across the $$3\text{ µF}$$ capacitor.
- Equivalent $$1/C_s = 1/2+1/3+1/6=1$$ ⇒ $$C_s=1\text{ µF}$$.
- Total charge $$Q= C_s V = 1\text{ µF}\times120\text{ V}=120\text{ µC}$$.
- Voltage on $$3\text{ µF}: V = Q/C = 120/3 = 40\text{ V}$$.
Answer: $$C_{eq}=1\text{ µF},\; Q=120\text{ µC},\; V_{3\text{ µF}}=40\text{ V}$$
Shortcuts and Special Cases
- Two equal capacitors: series gives $$C/2$$, parallel gives $$2C$$ – keep in mind.
- Star-delta transformation for three equal $$C$$ gives equivalent $$C$$ of $$C$$ – rare but fast.
Watch Out
- Energy lost when two charged capacitors are connected is due to redistribution; examiner expects you to calculate it.
- Series connection potential can exceed supply on an individual capacitor – watch dielectric breakdown limits.
Energy Stored & Energy Density Formulas
Energy formulas appear in every variant: moving plates, inserting dielectrics, or explosion of capacitors. You must juggle three interchangeable forms and know when to use volume density.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| $$U = \frac12 CV^{2} = \frac12 QV = \frac{Q^{2}}{2C}$$ | Total stored energy | Static case |
| Energy density $$u = \frac12 \varepsilon_0 E^{2}$$ | Energy per volume in field | Vacuum/air |
| With dielectric: $$u = \frac12 \kappa\varepsilon_0 E^{2}$$ | Energy density in dielectric | Linear isotropic $$\kappa$$ |
| Force on plate: $$F = \frac12 \varepsilon_0 \frac{A V^{2}}{d^{2}}$$ | Pull of one plate on other | Parallel plate, constant $$V$$ |
| Pressure $$P = \frac12 \varepsilon_0 E^{2}$$ | Electrostatic pressure on surface | — |
Worked Example
A parallel-plate capacitor has $$C=5\text{ µF}$$ charged to $$400\text{ V}$$. If the plates are pulled apart while disconnected so that capacitance drops to $$2\text{ µF}$$, find the new voltage and energy change.
- Charge remains constant $$Q = CV = 5\times10^{-6}\times400 = 2.0\times10^{-3}\text{ C}$$.
- New voltage $$V' = Q/C' = 2.0\times10^{-3}/(2\times10^{-6}) = 1000\text{ V}$$.
- Initial energy $$U_i = \frac12 CV^{2}=0.5\times5\times10^{-6}\times160000=0.4\text{ J}$$.
- Final energy $$U_f = Q^{2}/(2C') = (2\times10^{-3})^{2}/(2\times2\times10^{-6})=1.0\text{ J}$$.
- Energy increases by $$0.6\text{ J}$$ – supplied by mechanical work.
Answer: $$V'=1.0\text{ kV},\; \Delta U = +0.6\text{ J}$$
Shortcuts and Special Cases
- Constant $$V$$: inserting dielectric lowers energy by factor $$1/\kappa$$.
- Constant $$Q$$: inserting dielectric lowers voltage by $$1/\kappa$$ and energy by same factor.
Watch Out
- State whether battery connected (constant $$V$$) or isolated (constant $$Q$$) – most common trap.
- Force formula halves because energy only on one side; many students miss 1/2.
Dielectrics in Capacitors Formulas
Dielectrics pop up as slabs fully or partially introduced, affecting field, capacitance and energy. Examiner tests boundary conditions and concept of induced surface charge.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Capacitance with dielectric slab thickness $$t$$: $$C = \frac{\varepsilon_0 A}{d-t + t/\kappa}$$ | Effective capacitance | Slab fills area $$A$$, parallel plates |
| Partial insertion length $$x$$ (no battery): $$F = \frac12 V^{2} \frac{dC}{dx}$$ | Force pulling slab in | Neglect fringing |
| Bound surface charge: $$\sigma_b = P\cdot\hat n$$, $$P=(\kappa-1)\varepsilon_0 E$$ | Polarisation charge density | Linear dielectric |
| Field inside dielectric: $$E = E_{0}/\kappa$$ (if battery disconnected) | Reduced internal field | Uniform slab, plates isolated |
Worked Example
Between plates separated by $$4\text{ mm}$$ (area $$200\text{ cm}^2$$) a glass slab $$\kappa=5$$ of thickness $$1\text{ mm}$$ is inserted fully. Find new capacitance.
- Convert: $$A=200\text{ cm}^{2}=0.02\text{ m}^2,\, d=0.004\text{ m},\, t=0.001\text{ m}.$$
- Effective gap $$d_{eq}=d-t+t/\kappa = 0.004-0.001+0.001/5 = 0.003+0.0002=0.0032\text{ m}.$$
- Capacitance $$C = \varepsilon_0 A/d_{eq}=8.85\times10^{-12}\times0.02/0.0032 = 5.53\times10^{-11}\text{ F}=55.3\text{ pF}.$$
Answer: $$55\text{ pF}$$
Shortcuts and Special Cases
- Thin dielectric $$t\ll d$$: fractional change $$\approx(\kappa-1)t/d$$ – handy for percentage questions.
- Capacitor partially filled lengthwise acts like two capacitors in parallel (slab & air). Widthwise insertion – series.
Watch Out
- Which dimension slab moved? Mix-ups lead to wrong parallel/series treatment.
- Area conversion cm² to m²: divide by 10 000.
Charging & Discharging of a Capacitor through a Resistor Formulas
Though strictly current electricity, JEE inevitably tags one RC time constant problem in this chapter. You must know exponential laws and be able to find time for a fraction of charge.
Formulas
| Formula | What it gives you | Valid when |
|---|---|---|
| Charging: $$Q(t)=CV_0(1-e^{-t/\tau})$$ | Charge at time $$t$$ | Switch closed at $$t=0$$ |
| Discharging: $$Q(t)=Q_0 e^{-t/\tau}$$ | Remaining charge | Isolated RC |
| Voltage across capacitor: same form as $$Q/C$$ | Instantaneous voltage | — |
| Current: $$I(t) = (V_0/R)e^{-t/\tau}$$ (charging) | Decay of current | — |
| Time constant: $$\tau = RC$$ | Characteristic time | Linear, constant $$R$$ |
Worked Example
A $$10\text{ µF}$$ capacitor is charged through a $$1\text{ M}\Omega$$ resistor from a $$12\text{ V}$$ battery. How long until the capacitor reaches $$9\text{ V}$$?
- Time constant $$\tau=RC=1\times10^{6}\times10^{-5}=10\text{ s}$$.
- Charging formula $$V=V_0(1-e^{-t/\tau})$$.
- $$9/12 = 1-e^{-t/10}$$ ⇒ $$e^{-t/10}=0.25$$.
- Take ln: $$-t/10 = \ln0.25 = -1.386$$ ⇒ $$t=13.86\text{ s}$$.
Answer: $$14\text{ s (approx)}$$
Shortcuts and Special Cases
| Fraction reached | t / τ (charging) | t / τ (discharging) |
|---|---|---|
| 63 % | 1 | — |
| 90 % | 2.3 | — |
| 37 % | — | 1 |
| 10 % | — | 2.3 |
Watch Out
- Capacitance in microfarad, resistance in mega-ohm gives τ directly in seconds (10⁶×10⁻⁶=1).
- Don’t mix discharge and charge curves – one minus exponential vs plain exponential.
Formulas That Look Alike
| Confusing Pair | Context 1 | Context 2 | How to decide |
|---|---|---|---|
| $$V=kQ/r$$ vs $$E=kQ/r^{2}$$ | Scalar potential | Vector field | If question talks about work or energy, use $$V$$. |
| $$C=\varepsilon_0 A/d$$ vs $$C=4\pi\varepsilon_0 R$$ | Parallel plates | Isolated sphere | Presence of two conductors? Use plate. |
| Energy density $$\frac12\varepsilon_0E^{2}$$ vs Pressure $$\frac12\varepsilon_0E^{2}$$ | Energy per volume | Force per area | Ask yourself: is it asking for force? |
| Series $$1/C$$ sum vs Resistor series $$R$$ sum | Capacitors | Resistors | Capacitor shares charge, resistor shares current. |
Common Mistakes to Avoid When Using Electric Potential & Capacitance Formulas for JEE
- Forgetting to convert centimetres, millimetres, micro-coulombs into SI before plugging $$k=9\times10^{9}$$.
- Writing natural log but calculating with base-10 log in coaxial capacitor questions.
- Using constant $$V$$ formula when battery is disconnected and vice versa.
- Missing the 1/2 factor in energy or force expressions.
- Adding potentials vectorially – potential is a scalar, always algebraic sum with sign.
- Using series rule for capacitors where plates share common potential (actually parallel).
- Expecting zero potential inside conducting shell to mean zero everywhere inside cavity with enclosed charge – not true.
Quick Revision of Electric Potential & Capacitance Formulas for JEE 2027
- Point charge potential: $$V=kq/r$$.
- Parallel-plate capacitor: $$C=\varepsilon_0A/d$$.
- Isolated sphere: $$C=4\pi\varepsilon_0R$$.
- Energy stored: $$U=\frac12CV^{2}=\frac{Q^{2}}{2C}$$.
- Series: $$1/C_s=\sum1/C_i$$; Parallel: $$C_p=\sum C_i$$.
- Capacitance with dielectric slab: $$C=\varepsilon_0A/(d-t+t/\kappa)$$.
- RC time constant: $$\tau=RC$$, 63 % charge in one τ.
- Energy density in field: $$u=\frac12\varepsilon_0E^{2}$$.
Electric Potential and Capacitance Formulas for JEE 2027: Conclusion
Electric Potential and Capacitance formulas for JEE 2027 help you approach questions on charge distributions, equipotential surfaces, capacitor networks, dielectrics and stored energy. Use this sheet to revise the key relations alongside their conditions and worked examples. Understanding charge signs, capacitor geometry and the difference between constant charge and constant voltage will help you apply the formulas correctly.
Strengthen your preparation by solving JEE Electric Potential and Capacitance PYQs, practising JEE Capacitance Questions and reviewing JEE Main previous papers. Combine this formula sheet with your JEE study material and Physics mock tests to identify gaps in understanding. Regular revision, careful unit conversions and reattempting incorrect questions can improve your speed and accuracy for JEE Main and Advanced.
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