Question

A Rod of length has a total charge uniformly distributed along its length. The rod lies along the axis with its center at the origin.

a. Find an expression for the electric potential as a function of position along the axis.
b. Show that the result obtained in Part (a) reduces to for Explain why this result is expected

Answer 1

Answer:

a) dE={KdQdX}{r^{2} } *cosT

b) E = (2Q)/4eL^2)

Explanation:

a) The charge of element is:

dQ = Q/L

Where the y-axis elements will cancel, thus:

$dE=\frac{KdQdX}{r^{2} } *cosT$

where dX is the length element = RdT and dT is the angle element, K = 1/(4pe)

b) The angle from x-axis is:

$dE=\frac{KdQ*RdX}{r^{2} } *cosT$

We integrate the expression from T to p/2 and multiply by 2:

$E=\frac{2KdQ}{R} =\frac{2KQ}{LR} \frac{2KQ_{p} }{L^{2} }$

Replacing the k value:

$E=\frac{2Q}{4eL^{2} }$