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

Coulombs force between two-point charges

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

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

Electric field

3. Electric field Intensity

E=F/q0\vec{E}=\vec{F}/q_{0}

Where F is the force on q due to E

4. Electrostatic Energy

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

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

Electric dipole moment

p=qd\vec{p}=q\vec{d}

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

Er=14πϵ02pcosθr3E_{r}=\frac{1}{4\pi \epsilon _{0}}\frac{2pcos\theta }{r^{3}} Eθ=14πϵ0pcosθr3E_{\theta }=\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

U=p.EU=-\vec{p}.\vec{E}