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# Class_3 - 3 1 4 n n n n q-=-∑ r r E r r 1 4 n n n q =-∑...

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PHYS809 Class 3 Notes Electric field and potential for a distribution of discrete charges The electric field at position r from a single charge q at the origin is 1 2 ˆ, q k r = E r where ˆ r = r r is a unit vector in the radial direction. In SI units, ( ) 1 1 0 4 . k πε - = The permittivity of free space is 12 1 0 8.85410 Farad m . ε - - We can find the divergence of E by applying the divergence theorem to a spherical volume of radius a centered on the charge. Since the field is spherically symmetric, 2 2 0 0 4 . 4 V S q q dV d a a π ∇⋅ = = = E E S Since this is independent of a , we can conclude that ( ) 0 . q δ ∇⋅ = E r Due to the spherical symmetry, the electrostatic potential is a function of r alone. Hence . r d E dr φ = - Fixing the constant of integration by setting the potential to be zero at infinity, we get ( ) 0 . 4 q = r r Also, because , = -∇ E the potential satisfies ( ) 2 0 . q = - r For a set of discrete charges q 1 , q 2 , … at positions r 1 , r 2 , … the above expressions generalize to

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Unformatted text preview: 3 1 , 4 n n n n q-=-∑ r r E r r ( ) 1 , 4 n n n q =-∑ r r r ( ) ( ) 2 1 , 1 . n n n n n n q q δ ε φ ∇⋅ =-∇ = --∑ ∑ E r r r r Electric field and potential for a continuous distribution of charges By dividing a continuous distribution of charges into small pieces and approximating each piece by a discrete charge, and then taking the limit in which the size of each piece goes to zero, it can be shown that for a charge distribution ρ ( r ) in volume V , the electric field and potential at position R are given by ( ) ( ) 3 3 1 , 4 V d ρ πε-=-∫ R r E R r r R r ( ) ( ) 3 1 . 4 V d =-∫ r R r R r Also 2 , . ∇⋅ = ∇ = -E The expressions for the electrical field are both referred to as Gauss’s law....
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Class_3 - 3 1 4 n n n n q-=-∑ r r E r r 1 4 n n n q =-∑...

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