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| Application of Gauss' Theorem |
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| Gauss' theorem can be used to calculate the electric intensity due to |
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 an infinitely long straight charged wire |
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 a uniformly charged infinite plane sheet |
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 a uniformly charged thin spherical shell |
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| Consider a thin long charged wire. Let the charge per unit length of the wire be l. To calculate the field at P we consider a Gaussian surface with wire as axis, radius r and length l as shown in the figure. This Gaussian surface that is, the cylinder is closed at each end by planes normal to the axis. |
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| The electric lines of force are parallel to the end faces of the cylinder and hence the component of the field along the normal to the end faces is zero. |
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| The field is radial everywhere and hence the electric flux crosses only through the curved surface of the cylinder. |
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| If E is the electric field intensity at P, then the electric flux through the Gaussian surface is E x 2prl(2prl is the surface area of the curved part) |
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| The charge enclosed by the Gaussian surface is ll |
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 is
directed radially outwards if q is positive and radially inwards if q is negative. |
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| Note: |
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| SI unit of l is C/m. |
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| Let s be the uniform surface charge density of an infinite plane sheet. |
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| If x-axis is taken normal to the given plane then the electric field will not depend on y and z axes. |
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| The Gaussian surface will be a parallelepiped of cross sectional area A. only the two faces 1 and 2 will contribute to the flux whereas the other two faces do not contribute to the total flux as the electric field lines are parallel to them. |
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| Flux through the Gaussian surface = 2EA |
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| (The factor 2 appears as the total flux is due to two faces of parallelepiped which are normal to electric field |
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| SI unit of s is C/m2 |
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| Consider a hollow conducting sphere of radius R with its centre at O. let s be its surface density. The field at any point P, outside or inside depends upon the distance from the centre of the spherical shell. Let the distance between the centre of the spherical shell and the point be r. |
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| Case 1 |
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| Field Inside a Hollow Conducting Sphere |
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| Consider a thin hollow conducting sphere with radius R. Let q be the charge on this sphere. To find the field at a point P, draw a gaussian surface (dotted circle) of radius r. Since, this surface does not enclose any charge, we have |
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| Case 2 |
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| Field on the surface of the shell |
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| When point P lies on the surface of the shell or sphere, r = R so |
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| Case 3 |
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| Field at a Point Outside the Spherical Shell |
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| Let the point P be outside the spherical shell. At points outside the sphere the electric field is radial every where because of spherical symmetry. |
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where is the unit
vector.is directed outwards if the charge is positive and is directed inwards if the charge is negative. |
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| The electric field in terms of surface charge density |
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| Case I |
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| Case II |
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| Case III |
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