Question

A 12-cm-long thin rod has the nonuniform charge density λ(x)=(2.0 nC/cm)e−|x|/(6.0 cm), where x is measured from the center of the rod. What is the total charge on the rod? Hint: This exercise requires an integration. Think about how to handle the absolute value sign

Answer 1

Answer:

the total charge is

$Q=24(1-\exp(-1))nC\approx15.171nC$

Step-by-step explanation:

Since x is measured from the center, that means that x=0 is the center so the edges of the rod correspond to x=-6 and x=6. that meas that the total charge can be calculated as

$Q=\int^{6}_{-6}2\exp\left(\frac{-|x|}{6}\right)dx$

separating the integral from -6 to 0 and from 0 to 6, taking into account that |x|=-x for x<0 and |x|=x for x >=0, we get$Q=\int^{0}_{-6}2\exp\left(\frac{x}{6}\right)dx+\int^{6}_{0}2\exp\left(\frac{-x}{6}\right)dx$

using the substitution x=-u in the first integral we get$\int^{0}_{-6}2\exp\left(\frac{x}{6}\right)dx=-\int^{0}_{6}2\exp\left(\frac{-u}{6}\right)du=\int^{6}_{0}2\exp\left(\frac{-u}{6}\right)du$

which is the same as the first integral. Thus, the total charge is given by

$Q=2\int^{6}_{0}2\exp\left(\frac{-x}{6}\right)dx$

integrating we get

$Q=4(-6\exp\left(\frac{-x}{6}\right))\big|^{6}_{0}=-24(\exp(-6/6)-\exp(0))=24(1-\exp(-1))$