gauss law and application Arun kumar
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This document provides an overview of Gauss's law and its applications. It begins with definitions of electric flux and how to calculate flux through surfaces. It then introduces Gauss's law, which relates the electric flux through a closed surface to the enclosed charge. Examples are provided to demonstrate how to use Gauss's law to determine electric fields for symmetric charge distributions. The document also discusses how the electric field is zero inside conductors in electrostatic equilibrium and how Gauss's law can be used to show this. Worked examples further illustrate applying Gauss's law.
gauss law and application Arun kumar
- 2. Presented BY NAME : ARUN KUMAR CLASS : BSC IV SEM (C.S.) COLLEGE : RAI SAHEB BHANWAR SINGH COLLEGE NASRULLAGANJ SUBMITTED TO: GYAN RAO DHOTE RAI SAHEB BHANWAR SINGH COLLEGE NASRULLAGANJ 2
- 3. RAI SAHEB BHANWAR SINGH COLLEGE NASRULLAGANJ 3
- 4. Gauss’s Law
- 5. Basic Concepts Electric Flux Gauss’s Law Applications of Gauss’s Law Conductors in Equilibrium Physics 24-Winter 2003-L03 5
- 6. Electric Flux Physics 24-Winter 2003-L03 6 The electric flux, FE, through a surface is defined as the scalar product of E and A, FE = EA. A is a vector perpendicular to the surface with a magnitude equal to the surface area. This is true for a uniform electric field. A = A cos so FE = EA = EA cos FE = EA
- 7. Electric Flux Continued Physics 24-Winter 2003-L03 7 What about the case when the electric field is not uniform and the surface is not flat? Then we divide the surface into small elements and add the flux through each. E i i i E d A E d AF r rr r
- 8. Worked Example 1 Physics 24-Winter 2003-L03 8 Compute the electric flux through a cylinder with an axis parallel to the electric field direction. E The flux through the curved surface is zero since E is perpendicular to dA there. For the ends, the surfaces are perpendicular to E, and E and A are parallel. Thus the flux through the left end (into the cylinder) is –EA, while the flux through right end (out of the cylinder) is +EA. Hence the net flux through the cylinder is zero. A
- 9. Gauss’s Law Physics 24-Winter 2003-L03 9 Gauss’s Law relates the electric flux through a closed surface with the charge Qin inside that surface. 0 F rr Ñ in E Q E dA This is a useful tool for simply determining the electric field, but only for certain situations where the charge distribution is either rather simple or possesses a high degree of symmetry.
- 10. Problem Solving Strategies for Gauss’s Law Select a Gaussian surface with symmetry that matches the charge distribution Draw the Gaussian surface so that the electric field is either constant or zero at all points on the Gaussian surface Use symmetry to determine the direction of E on the Gaussian surface Evaluate the surface integral (electric flux) Determine the charge inside the Gaussian surface Solve for E Physics 24-Winter 2003-L03 10
- 11. Conductors in Electrostatic Equilibrium The electric field is zero everywhere inside the conductor Any net charge resides on the conductor’s surface The electric field just outside a charged conductor is perpendicular to the conductor’s surface Physics 24-Winter 2003-L03 11 By electrostatic equilibrium we mean a situation where there is no net motion of charge within the conductor
- 12. Conductors in Electrostatic Equilibrium Physics 24-Winter 2003-L03 12 Why is this so? If there was a field in the conductor the charges would accelerate under the action of the field. The electric field is zero everywhere inside the conductor ++++++++++++ --------------------- Ein E E The charges in the conductor move creating an internal electric field that cancels the applied field on the inside of the conductor
- 13. Worked Example 4 Physics 24-Winter 2003-L03 13 Any net charge on an isolated conductor must reside on its surface and the electric field just outside a charged conductor is perpendicular to its surface (and has magnitude σ/ε0). Use Gauss’s law to show this. For an arbitrarily shaped conductor we can draw a Gaussian surface inside the conductor. Since we have shown that the electric field inside an isolated conductor is zero, the field at every point on the Gaussian surface must be zero. From Gauss’s law we then conclude that the net charge inside the Gaussian surface is zero. Since the surface can be made arbitrarily close to the surface of the conductor, any net charge must reside on the conductor’s surface. 0 rr Ñ inQ E dA
- 14. Worked Example 4 cont’d Physics 24-Winter 2003-L03 14 We can also use Gauss’s law to determine the electric field just outside the surface of a charged conductor. Assume the surface charge density is σ. Since the field inside the conductor is zero there is no flux through the face of the cylinder inside the conductor. If E had a component along the surface of the conductor then the free charges would move under the action of the field creating surface currents. Thus E is perpendicular to the conductor’s surface, and the flux through the cylindrical surface must be zero. Consequently the net flux through the cylinder is EA and Gauss’s law gives: 0 0 0 orin E Q A EA E F
- 15. Worked Example 5 Physics 24-Winter 2003-L03 15 A conducting spherical shell of inner radius a and outer radius b with a net charge -Q is centered on point charge +2Q. Use Gauss’s law to find the electric field everywhere, and to determine the charge distribution on the spherical shell. a b -Q First find the field for 0 < r < a This is the same as Ex. 2 and is the field due to a point charge with charge +2Q. 2 2 e Q E k r Now find the field for a < r < b The field must be zero inside a conductor in equilibrium. Thus from Gauss’s law Qin is zero. There is a + 2Q from the point charge so we must have Qa = -2Q on the inner surface of the spherical shell. Since the net charge on the shell is -Q we can get the charge on the outer surface from Qnet = Qa + Qb. Qb= Qnet - Qa = -Q - (-2Q) = + Q. +2Q
- 16. Worked Example 5 cont’d Physics 24-Winter 2003-L03 16 a b -Q +2Q Find the field for r > b From the symmetry of the problem, the field in this region is radial and everywhere perpendicular to the spherical Gaussian surface. Furthermore, the field has the same value at every point on the Gaussian surface so the solution then proceeds exactly as in Ex. 2, but Qin=2Q-Q. 2 4 rr Ñ Ñ ÑE dA E dA E dA E r Gauss’s law now gives: 2 2 2 0 0 0 0 2 1 4 or 4 in e Q Q Q Q Q Q E r E k r r