Ch3+Gauss+Law - GAUSS LAW ALI ALZAHRANI ELECTRIC FLUX...

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G AUSS L AW A LI A LZAHRANI
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Electric flux ° is the drift of the electric field through a certain surface. Mathematically it is defined as closed integral of the scalar product of electric field and surface area ± = ² ? . ?³ ´ The enclosed surface is called Gaussian Surface. For uniform area, the above integral tends to be ± = ? . ³ ´ The electric flux through a Gaussian surface is proportional to the electric field lines passing through that surface. Zero flux occurs when the surface is normal to the electric field lines. E LECTRIC F LUX
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Gauss law states that the electric flux over a closed surface is proportional to the total electric charge contained in the surface, ? °±² . ³ = ? °±² ´ 0 The above equation suggests that the electric flux does not depend on the shape of the surface. The Gauss law can be written as ? . ?µ = ? °±² ´ 0 The unit of electric flux is N. m 2 /C.
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Suppose we have a point charge q, the electric field at any point r from the charge can be evaluated from Gauss law E . dA = q enc ε 0 First take a Gaussian surface ‘sphere’ of radius r Using the above formula with q enc = ? , we get ?° = q enc ε 0 = ? ε 0 ±±±±→ ±±? 4²? 2 = ? ε 0 ? = ? 4²ε 0 ? 2 = ³±? ? 2 r Gaussian Surface q
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Suppose we have a line of charge with λ , the electric field at any point r from the line can be evaluated from Gauss law E . dA = q enc ε 0 First take a Gaussian surface ‘cylinder’ of radius r and length L Using the above formula with q enc = °L , we get ?± = q enc ε 0 = °L ε 0 ²²²²→ ²²? 2³?´ = °L ε 0 ? = ° 2³ε 0 ? λ r L Gaussian Surface
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Suppose we have an infinite conducting sheet having surface charge ° , the electric field at any point r from the sheet can be evaluated from Gauss law E . dA = q enc ε 0 First take a Gaussian surface ‘cylinder’ of cross-section area A Using the above formula with q enc = ° A, we get ?± = q enc ε 0 = ° A ε 0 ²²²²²²²²²²²²²²²²²→ ²²²²²²²²? = ° ε 0 If the sheet is non-conducting, the electric field is ?± + ?± = q enc ε 0 = ° A ε 0 ²²²²²²→ ²²²²²²²²? = ° 0 Conducting sheet Non-conducting sheet
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Suppose we have a conducting sphere of radius R and charge q, the electric field at any point r from the center can be evaluated from Gauss law E . dA = q enc ε 0 We have two region to find the electric field at, inside and outside the sphere. Let us determine the field inside the sphere. Take a Gaussian surface ‘sphere’ of radius r. For any conducting sphere, the charge is distributed over its surface and no charge is enclosed within it. Therefore using the above formula with q enc = ± , we get ? = ± R r q
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Now let us determine the field outside the sphere. Take a Gaussian surface ‘sphere’ of radius r
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This note was uploaded on 02/18/2012 for the course PHYSICS 202 taught by Professor 1 during the Spring '12 term at King Abdulaziz University.

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Ch3+Gauss+Law - GAUSS LAW ALI ALZAHRANI ELECTRIC FLUX...

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