Derivation of electric field due to a infinite charge along line
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wire
E.ds= qnet/E°------1
qnet= ∆l (∆=charge per unit length)
Ed's=For component 1,2,3
E.ds(|||) = E.dscos 0°
=E intg. ds
=E(2πrl)
So putting in Eqn 1 we get
E(2πrl)=∆l/E° (ds= 2πrl)
E=1/2πE° × ∆/r
E=1/4πE° × 2∆/r
E.ds= qnet/E°------1
qnet= ∆l (∆=charge per unit length)
Ed's=For component 1,2,3
E.ds(|||) = E.dscos 0°
=E intg. ds
=E(2πrl)
So putting in Eqn 1 we get
E(2πrl)=∆l/E° (ds= 2πrl)
E=1/2πE° × ∆/r
E=1/4πE° × 2∆/r
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The electric field of an infinite line charge with a uniform linear charge density can be obtained by a using Gauss' law. Considering a Gaussian surface in the form of a cylinder at radius r, the electric field has the same magnitude at every point of the cylinder and is directed outward.
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