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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Important Formulas » JEE Physics Important Formulas Part 1

JEE Physics Important Formulas Part 1

In this article we will go through physics quick formula revision for JEE. Track down the important formulas of Uniform Circular Motion, Alternating Current, Ampere’s Circuital Law and Capacitance.

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Uniform Circular Motion Formula

The formula for uniform circular motion are as stated below
Description Formula
The formula for Angular Distance is Δθ = ω Δt, Where t is time, ω is angular speed and θ is angular distance.
The formula for linear velocity is given by v = Rω Where speed and R is radius and ω is angular speed.
The formula for Centripetal Acceleration is given by Ac = v2/R, Where R is the radius and v is the velocity. Ac = ω2R Where R is the radius and ω is angular speed Ac = 4π2v2R Where R is the radius and ν is the frequency
Average Angular Velocity \omega_{av}=\frac{\theta_2-\theta_1}{t_2-t_1}=∆θ∆t
Instantaneous angular Velocity ω=dθ/dt
Average Angular acceleration \alpha_{av}=\frac{\omega_2-\omega_1}{t_2-t_1}=∆ω∆t
Instantaneous angular acceleration α=dωdt=ωdωdθ
Relation between speed and angular velocity v=rω  and v=r
Tangential acceleration a_t=\frac{dV}{dt}=r\frac{d\omega}{dt}=\omega\frac{dr}{dt}
Radial or normal or centripetal acceleration a_r=\frac{V^2}{r}=\omega^2r
Angular Acceleration \vec{\alpha}=\frac{d\vec{\omega}}{dt}\ (\ Non-uniform\ motion)
Normal reaction of road on a concave bridge N=mg\ cos\theta+\frac{mv^2}{r}
Normal reaction on a convex bridge N=mg\ cos\theta+\frac{mv^2}{r}
Skidding of vehicle on a level road V_{safe}\le\sqrt{\mu gr}
Skidding of an object on a rotating platform
\omega_{max}=\sqrt{\frac{\mu g}{r}}
Bending of Cyclist tan\ \theta=\frac{v^2}{rg}\
Banking of road without friction tan\ \theta=\frac{v^2}{rg}\
Banking of Road with friction \frac{V^2}{rg}=\frac{\mu+tan\theta}{1-\mu\ tan\ \theta\ }
Maximum also minimum safe speed on a banked frictional road \frac{V^2}{rg}=\frac{\mu+tan\theta}{1-\mu\ tan\ \theta\ }

Alternating Current Formula

The formula for alternating current are as stated below
AC and DC current
A current that changes its direction periodically is called alternating current (AC). If a current maintains its direction constant it is called direct current (DC).
Root Mean square Value Root mean square of a function from  is defined as f_{rms}=\sqrt{\frac{\int_{t_i}^{t_2}{f(t)}^2dt}{t_2-t_1}}
Power consumption in AC Circuit Average power consumed in a cycle = \frac{1}{T}\int_{0}^{\frac{2\pi}{\omega}}\frac{Pdt}{\frac{2\pi}{\omega}}=\frac{1}{2}V_mI_mcos\phi\ \ \ =\frac{V_m}{\sqrt2}.\frac{I_m}{\sqrt2}.cos\phi=V_{rms}I_{rms}cos\phi cos\phi Is known as the Power Factor.
Impedance z=\frac{V_m}{I_m}=\frac{V_{rms}}{I_{rms}}
Purely Resistive Circuit
I=\frac{V_S}{R}=\frac{V_msin\omega t}{R}=I_msin\omega t I_m=\frac{V_m}{R} I_{rms}=\frac{V_{rms}}{R} < p >=V_{rms}I_{rms}cos\phi=\frac{V_{rms}^2}{R}
Purely Capacitive Circuit I=\frac{\frac{V_m}{1}}{\omega c}cos\omega t =\frac{V_m}{X_c}cos\ \omega t=I_mcos\omega t\ X_c=\frac{1}{\omega C\ } And is called capacitive reactance. lc Leads by Vc  by Π/2 , Diagrammatically it is represented as Since, \phi={90}^0, =V_{rms}I_{rms}cos\phi=0

Ampere’s Circular Law

The formula for Ampere’s circuital law  are as stated below
Description Formula
Ampere’s circuital law \int_{\ }^{\ }B.dl=μ°I Here μ°= permeability of free space=4Π×10-15NA-2 B = Magnetic field I = enclosed electric current by the path
Ampere’s law (integral form) \int_{\ }^{\ }B.ds=μ°Ienclosed Ienclosed= enclosed current by the surface
Field of a current-carrying wire: B=μ°/I2πr
Field of a solenoid BL=°NI Here N: number of turns in the solenoid
Field inside a thick wire \int B.ds=μ°I And B=μ°I.r2/πR2
Field of the toroid B=μ°NI/2πr
Force between two parallel current carrying wires F_\frac{A}{B}=μ°IAIB2πr IA,IB= Current carrying by wires A and B

Capacitance formula

The formula for capacitance are as stated below
Description Formula
Capacitance of a parallel plate capacitor in terms of charge and potential difference C=Q/V Here, C is the capacitance of the capacitor, Q is the charge stored and V is the potential difference between the plates.
Capacitance of a parallel plate capacitor in terms of surface area and distance between the plates C=\frac{\varepsilon_0A}{d} Here,  is the permittivity of free space and its value is 8.854×10-12m-3 Kg-1 s4 A2 is the surface area of the plates and d is the distance between the plates.
Capacitance of a spherical capacitor derivation To find the formula for capacitance of a spherical capacitor we will use the gauss’s law. Let the charge on the spherical surface be , the radius of smaller sphere be  and radius of the bigger sphere be . Using gauss’s law, we can write: \oint_{\ }^{\ }\ \vec{E}\cdot d\vec{A}=\frac{Q}{\varepsilon_0} E\left(4\pi r^2\right)=\frac{Q}{\varepsilon_0} E=\frac{Q}{4\pi\varepsilon_0r^2} V=\frac{Q}{4\pi\varepsilon_0r}
The potential difference between the plates V_{ab}=V_a-V_b=\frac{Q}{4\pi\varepsilon_0}\left(\frac{1}{r_a}-\frac{1}{r_b}\right) =\frac{Q}{4\pi\varepsilon_0}\frac{r_b-r_a}{r_ar_b} Therefore, the capacitance will be: C=\frac{Q}{V_{ab}}=\ 4\pi\varepsilon_0\frac{r_ar_b}{r_b-r_a}
Energy stored in capacitor ● U=1/2 CV2 ● U= Q2/2c ● U= QV/2 Here, U is the energy, C is the capacitance, V is the potential difference and Q is the charge stored.
Energy density of capacitor Energy density=\frac{1}{2}\varepsilon_0\varepsilon_rE^2 In vacuum: Energy density= \frac{1}{2}\varepsilon_0E^2 Here, Eo is the permittivity of free space,  is the relative permittivity and E is the electric field.
Capacitance per unit length of a cylindrical capacitor Capacitance per unit length= 2ΠEo/In(b/a) Here, Eo is the permittivity of free space, b is the radius of outer cylinder and a is the radius of inner cylinder.
Electric field intensity The formula for electric field intensity between the plates is given as: E= σ/Eo=V/d Here, σ is the surface charge density, V is the potential difference and d is the distance between plates.
Redistribution of charge when two charged capacitors are connected in parallel Let us assume a capacitor with capacitance  with initial charge  and capacitor with capacitance  with initial charge . The final charge on capacitor with capacitance will be: Q’1=C1/C1+c2(Q1+Q2) final charge on capacitor with capacitance will be: Q’2=C1/C1+c2(Q1+Q2)
Equivalent capacitance when capacitors are connected in series \frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\ldots+\frac{1}{C_n} Here,  is the equivalent capacitance and  are the capacitance of the capacitors.
Equivalent capacitance of the capacitors connected in parallel C_{eq}=C_1+C_2+C_3+\ldots C_n
Charging of capacitor q=q_0\left(1-e^{-\frac{t}{\tau}}\right) Here, q is the charge on the capacitor at time t,  is the time constant and  is the charge on the capacitor at steady state.
Discharging of capacitor q=q_0e^{-\frac{t}{\tau}} Here, q is the charge on the capacitor at time t,  is the time constant and  is the charge on the capacitor at steady state.

Also check:

  • JEE Physics Important Formulas Part 2
  • JEE Physics Important Formulas Part 3
  • JEE Physics Important Formulas Part 4
  • JEE Physics Important Formulas Part 5
  • JEE Physics Important Formulas Part 6
  • JEE Physics Important Formulas Part 7
  • JEE Physics Important Formulas Part 8
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JEE Advanced 2022 Syllabus JEE Preparation Books
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