Calculus Derivation:

Gauss’ Law Derivation:

- Since the line is infinite we could place the origin under the point of interest and turn the problem into two semi-infinite line charges that meet at the origin.
- Next consider two infinitesimal, differential charges
**dq**that are an equal distance from the origin. The x-components of the E-field of the two differential charges are in opposite directions at the any midway point and cancel each other out. - Tthe E-field at any distance from the wire will only point along the y-axis, perpendicular to the wire. More generally, the will be radial outwards or inwards towards the wire.
- The magnitude of the E-field can be found by determining the contributions from all
**dq's**betweem zero and infinity (i.e. by integrating the contribution of**dq)**and then doubling the value,**E = 2E**._{y} - Treating the differential charge
**dq**as a point charge, the y-component of the electric field is

- The differential charge can be written as
**dq = l dx**. Where**l**is charge per unit length which is constant.

- Integrating between zero and infinity,

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- Using an integral table,

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- Note that as
**x**approaches infinity the value of**y**can be neglected relative to**x**in the denominator of the first term so that this term approaches**x/x = 1.**

- We only solve the problem in one plane passing through the line charge - the xy-plane - but our results are true for any plane and any point. Changing
**y**to**r**to represent the distance of the point from the line charge, we obtain the generic answer.

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- What is amazing is that the E-field is not infinite everywhere in space since there is an infinite amount of charge on the infinite line charge. This happens because the E-field of a point charge drops off as one over distance squared.

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- First we pick a closed surface which is related to the symmetry of the E-field. Since the E-field is perpendicular to the line charge and has the same value at any fixed distance from the line charge.

- A cylinder centered on the axis of the line charge is the best choice since the E-field is constant on the outer wall of any cylinder and zero on the ends of the cylinder. This choice makes doing the integration to find the electric flux easy.

- Letting
**L**be the length of the cylinder,

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- The charge
**Q**can be found using the charge density_{inside}**l**and the length**L**,**Q**^{inside}**= l L.****E**in Gauss’ Law

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The same results as before.