Electric Field Near an Infinitely Long Line Charge
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,
- 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.
- 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.
Gauss’ Law Derivation:
- 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,
- The charge Qinside can be found using the charge density l and the length L , Qinside = l L. Solving for E in Gauss’ Law
The same results as before.