## Electric Charges and Fields

# Electric Flux, Electric Dipole and Gaus's Law

- Electric flux is defined as the dot product of Electric field intensity and Area \tt \phi = \overline{E} \cdot \overline{A}
- According to Gauss law the total flux linked with Gaussion surface is \tt \frac{1}{\varepsilon_{0}} times the net charge. \tt \phi = \oint \overline{E} \cdot \overline{d} A = \frac{q}{\varepsilon_{0}}
- If charge is kept at the centre of a cube then flux through cube = \tt \frac{Q}{\varepsilon_{0}} flux through face = \tt \frac{Q}{6 \varepsilon_{0}}
- If charge is kept at the centre of face then flux through cube = \tt \frac{Q}{2 \varepsilon_{0}}
- If charge is kept at the corner of a cube then flux from cube = \tt \frac{Q}{8 \varepsilon_{0}}. \phi_{face} = \frac{Q}{24 \varepsilon_{0}}
- Electric field at a distance ‘r’ from the axis of a long straight charged wire \tt E = \frac{\lambda}{2 \pi \varepsilon_{0} r}
- Electric potential at a distance r from the axis of a long straight charged wire \tt v = \frac{- \lambda}{2 \pi \varepsilon_{0}} \log_{e}^{r} + K
- Electric field due to a thin infinite sheet (Non conducting) \tt E = \frac{\sigma}{2 \varepsilon_{0}}
- Electrostatic potential due to a thin infinite sheet \tt V= - \frac{\sigma}{2 \varepsilon_{0}} r + K
- Electric field due to a thick infinite sheet (Non conducting) \tt E = \frac{\sigma}{\varepsilon_{0}}. and V = - \frac{\sigma}{\varepsilon_{0}} r + K
- Electric field due to a charged hollow sphere at a point out side the sphere \tt E = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{r}^{2} = \frac{\sigma R^{2}}{\varepsilon_{0} r^{2}} \left(r \gt R\right)
- Electric field due to a charged hollow sphere at a point on the sphere \tt E = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{r}^{2} = \frac{\sigma }{\varepsilon_{0}}
- Electric field due to a charged hollow sphere inside E = o.
- Electric potential due to a charged hollow sphere at a point outside the sphere \tt V = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{r} \left(r \gt R \right)
- Electric potential due to a charged hollow sphere at a point on the surface \tt V = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R} \left(r = R \right)
- V inside sphere = \tt \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R}
- Electric field ‘E’ due to a uniformly charged Non conducting sphere E out \tt E_{out} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{r^{2}} = \frac{\rho R^{3}}{3 \varepsilon_{0} r^{2}}
- E at the surface = \tt = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Q}{R^{2}} = \frac{\rho R}{3 \varepsilon_{0}}
- E inside = \tt = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Qr}{R^{3}} = \frac{\rho R}{3 \varepsilon_{0}}
- E, r graph for a hollow sphere

- E, r graph for a conducing sphere

- V, r graph for a hollow sphere
- Electric potential due to spherical charge distribution \tt V_{out} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r} \ \ V_{surface} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}
- Electric potential due to spherical charge distribution \tt V_{inside} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{2R} \left[3 - \frac{r^{2}}{R^{2}}\right]

V, r graph for spherical charge distribution

- Electric field at the axis of a circular uniformly charged ring \tt E = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{Qr}{\left(R^{2} + r^{2}\right)^{3/2}}
- Electric field at the centre of a uniformly charged ring E = O.
- Electric potential due to uniformly charged disc at a distance ‘r’ \tt V = \frac{\sigma}{2 \varepsilon_{0}} \left[\sqrt{R^{2} + r^{2}} -r \right]
- At centre r = o \tt \therefore V = \frac{\sigma R}{2 \varepsilon_{0}}
- Electrostatic potential energy of system of two charges q
_{1}and q_{2}separated by ‘d’. \tt U = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{d} - Electrostatic potential Energy of system of three charges q
_{1}, q_{2}and q_{3}\tt U = \frac{1}{4 \pi \varepsilon_{0}} \left[\frac{q_{1}q_{2}}{d_{1}} + \frac{q_{2}q_{3}}{d} +\frac{q_{3}q_{1}}{d}\right] - Two charges separated by a distance ‘d’ are left free. The gain in \tt KE = \frac{1}{8 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{d}
- Equation for two charges left free separated by ‘d’ \tt \frac{1}{2} mv^{2} + \frac{1}{2} mv^{2} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q_{1} q_{2}}{d}
- The amount of Energy gained by an electron when accelerated through a potential difference of one volt is Electron volt
- 1 ev = 1.6 × 10
^{−19}Joules. - Work done in rotating an electric dipole W = PE (cos θ
_{1}− cos θ_{2}) - When θ
_{1}= 0° θ_{2}= θ u = w = −PE cos θ - When θ
_{1}= 0 θ_{2}= 90° u = w = PE - For a metal conductor kept in Electric field the field inside the conductor is zero.
- The resultant electric field when a Dielectric is placed in an External field E = E
_{Applied field}– E_{induced}_{field} - Induced field Ei = E
_{Applied}\tt \left(1-\frac{1}{K}\right) (K = Dielectric const) - Induced charge on the surface \tt q_{i} = - q_{0} \left(1-\frac{1}{K}\right)
- Dielectric polarisation is defined as the dipole moment per unit volume of the dielectric.
- The relation between Electric displacement and polarisation. \tt \overline{D} = \varepsilon_{0} \overline{E} + \overline{P} (\tt \overline{P} = Dielectric polarisation, D = Electric displacement)
- The electric susceptibility (Χ
_{e}) and Dielectric polarisation are related as P = Χ_{e}E - The electric susceptibility Χ
_{e}= (K – 1) ε_{0} - Electric field intensity at the centre is zero (E = 0)

- Two equal and opposite charges separated by a fixed distance (2a) is called and electric dipole
- The electric dipole moment \tt \overline{P} = q \left(2\overline{a}\right)
- The direction of Electric dipole moment is from Negative charge to positive charge.
- Torque acting on Electric dipole Τ = PE sin θ.
- Electric field on the axial line of an electric dipole \tt E_{a} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2Pr}{\left(r^{2} - a^{2}\right)^{2}}
- Electric field on the Equatorial line of an electric dipole \tt E_{eq} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{P}{\left(r^{2} + a^{2}\right)^{3/2}}
- Electric field on the axial line of a short dipole \tt E_{a} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2P}{r^{3}}
- Electric field on the Equatorial line of a short dipole \tt E_{eq} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{P}{r^{3}}
- \tt \frac{E_{axial}}{E_{equitorial}} = 2
- Electric field due to short dipole at any point \tt E_{eq} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{P}{r^{3}} \sqrt{1 + 3 \cos^{2} \theta}
- n oil drops of radius ‘r’ and charge ‘q’ are merged to form a big drop of radius R charge Q. Then Radius of Big drop R = n
^{1/3}r - Charge on Big drop Q = nq
- Energy of Big drop U = n
^{2/3}. U_{small} - Electric field due to Big drop E = n
^{1/3}. E_{small} - Surface charge density of Big drop σ
_{Big }= n^{1/3}. σ_{small} - Potential of Big drop V
_{Big}= n^{2/3}V_{small} - Electric force between two dipoles placed coaxially \tt F = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{6 P_{1}P_{2}}{r4}
- Electric force between two dipoles placed perpendicular to each other \tt F = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{3 P_{1}P_{2}}{r4}
- Angular frequency (ω) of electric dipole in uniform electric field \tt \omega = \sqrt{\frac{PE}{I}}
- (I = moment of inertia)
- The electric dipole moment of the system is \tt \sqrt{3ql}

### Electric Dipole: View the Topic in this video From 41:56 To 54:10

### Gauss Law and Electric Flux

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1. Flux of electric field \tt \overrightarrow{E} through any area \tt \overrightarrow{A} is defined as

\tt \phi=E.A\cos \theta\ or\ \phi=\overrightarrow{E}.\overrightarrow{A}

2. In case of variable electric field or curved area. \tt \phi = \int \overrightarrow{E}.\overrightarrow{dA}

3. Electric Field Intensity on Axial Line

At the distance r from the centre of the electric dipole, \tt E=\frac{1}{4\pi \varepsilon_{0}}-\frac{2pr}{\left(r^2-a^2\right)^2}

4. Electric Field Intensity on Equatorial Line: At the point at a distance r from the centre of electric dipole, \tt E=\frac{1}{4\pi \varepsilon_{0}}\frac{p}{\left(r^2-a^2\right)^{3/2}}

5. The flux through an area element ΔS is

\tt \Delta \phi=E\cdot\Delta S=\frac{q}{4\pi\varepsilon_0r^2}\hat{r}\cdot\Delta S