Important Electrostatics Formulas For JEE

Electrostatics deals with the charges at rest. Charge of a material body or particle is the property due to which it produces and experiences electrical and magnetic effects. Some of the naturally occurring charged particles are electrons, protons etc. Unit of charge is Coulomb.

Electrostatics Formulas for JEE

1. Coulombs force between two-point charges


Coulomb's law


F=14πϵ0q1q2r2r^\vec{F}=\frac{1}{4\pi\epsilon _{0}}\frac{q_{1}q_{2}}{\left | \vec{r}\right |^{2}}\hat{r}

Here k= 1/4πε0 = 9 x 109 Nm2/C2

q1 and q2 are the charges separated by a distance r

2. Electric field

The electric field at a distance r from the charge q

E=14πϵ0qr2r^\vec{E}=\frac{1}{4\pi\epsilon _{0}}\frac{q}{{\left |\vec{r} \right |^{2}}}\hat{r}

3. Electric field Intensity


Where F is the force on the charge q due to the electric field E

4. Electrostatic Energy

U=14πϵ0q1q2r{U}=\frac{1}{4\pi\epsilon _{0}}\frac{q_{1}q_{2}}{r}

q1,q2 are the charges

r is the distance between the charges

5. Electric Potential

V=14πϵ0qrV=\frac{1}{4\pi\epsilon _{0}}\frac{q}{r} dV=E.rdV=-\vec{E}.\vec{r} V(r)=rE.drV(\vec{r})=-\int_{\infty }^{\vec{r}}\vec{E}.d{\vec{r}}

6. Electric dipole moment

It is the product of a charge (q) and the distance between the two charges (d)

Electric dipole moment



7. Potential of a dipole

Potential of a dipole


V=14πϵ0pcosθr2V=\frac{1}{4\pi \epsilon _{0}}\frac{pcos\theta }{r^{2}}

8. Field of a dipole

Field due to dipole


E+=14πϵ02pcosθr3E_{+}=\frac{1}{4\pi \epsilon _{0}}\frac{2pcos\theta }{r^{3}}


E=14πϵ0pcosθr3E=\frac{1}{4\pi \epsilon _{0}}\frac{pcos\theta }{r^{3}}

9. Torque on a dipole placed in the electric field

τ=p×E\vec{\tau }=\vec{p}\times \vec{E}

10. Potential energy of a dipole

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