Chapter 27 Electric Flux Lecture Notes PDF

## Chp # 27 Electric flux **Date: __________ 20** ### Guassian surface J ↳to find electric field intensity The number of electric fines of forces passing through (normally) a certain area is called electric flun. It is defined as scalar product of electric field intensity and vectol area. $Φ...

## Chp # 27 Electric flux **Date: __________ 20** ### Guassian surface J ↳to find electric field intensity The number of electric fines of forces passing through (normally) a certain area is called electric flun. It is defined as scalar product of electric field intensity and vectol area. $Φ = E.A$ The SI unit of electric flun is Nm²C^-1. ### Non uniform E If the electric field is not uniform, we divide the surface into small surfaces of area ΔA. Since the patches are very small we consider E to be constast for all the points. ### Discrete & continuous Achanges Area can be added: $Φ=EAcosθ$ ### **Case#1:** θ ≤ 90°  $E.ΔA → tue$ $Φε → tue$ ### **Case #02** θ > 90°  $E.ΔA → -ve$ $Φε → -ve$ direction of E = inward ### **Case III** θ=90° ![Diagram 3] (Not available) $E.ΔA → zero$ $Φε → zero$ ### Patch-wise total Φε **Provisional definition:** $Φε~=~\sum_{discrete} E.ΔA$ **Continuous sum - exact definition** $Φε = \int E.dA$ For a closed surface we draw cilde on integral: $Φε = \oint E.dA$ ### Gauss Law The total electric flun through any closed surface is *1 times charge* the total charge endosed by the surface: $Φe = \frac{q}{ε_ο}$ ε_ο = permittivity of free space = 8.854 x 10^-12 C²N^-1m^-2. where Φε in terms of E is: $Φε = \oint E.dA $ ### Lamparing (① & ②) ① $Φε = \oint E.dA$ ② $Φε = \frac{q}{ε_ο}$ → q = ε_ο * $\oint E.dA$ The circle on integral sign indicates that the integral is carried over a closed surface.  **At S1:** - +ve charge - integral gives +ve result - surface has +ve charge - Φε is +ve **At S2:** - -ve charge - integral gives -ve result - surface has -ve charge - Φε is -ve **At S3:** - no net charge - integral gives zero - surface contains two charges of equal magnitude but opposite signs. - flux is zero. ### Gauss law & Coulomb’s law We can deduce E'L by applying GOL:  Consider a spherical surface of radius r. E must be perpendicular to the surface so θ b/w E and da is zero. $q1 = E. \oint dA$ $q1 = E. \oint EdAcos(0)$ $q1 = E_o\oint EdA $ - ① $q1 = E_o\oint EdA$ - (1) where E = constant $q1 = E_o \oint dA$ - ② The total surface area of sphere is; $\oint dA = 4πr^2$ - (3) Put eq (3) in (2) $q1 = E_o . E(4πr^2)$ $⇒ Eo . E (4πr^2) = q1$ $⇒ E =\frac{1}{4πε_ο} \frac{q1}{r^2}$ ### 4 steps: 1. Charge Density 2. Gaussian Surface 3. Total flux by def: - Comparison with Φε 4. Flux by Gauss's law $Φ = B.A → flux by definition$ $Φ=q → flux by Gauss's law$ $E_o$ ### Applications of Gauss's law | 1D | 2D | 3D | | --- | --- | ---- | | wire | chips, rings | sphere | | | sheet of charges | solid | ### Infinite line of charge Consider a section of infinite line of charge having uniform charge density λ; λ= q/L