Answer:
the total charge is
[tex]Q=24(1-\exp(-1))nC\approx15.171nC[/tex]
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
[tex]Q=\int^{6}_{-6}2\exp\left(\frac{-|x|}{6}\right)dx[/tex]
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[tex]Q=\int^{0}_{-6}2\exp\left(\frac{x}{6}\right)dx+\int^{6}_{0}2\exp\left(\frac{-x}{6}\right)dx[/tex]
using the substitution x=-u in the first integral we get[tex]\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[/tex]
which is the same as the first integral. Thus, the total charge is given by
[tex]Q=2\int^{6}_{0}2\exp\left(\frac{-x}{6}\right)dx[/tex]
integrating we get
[tex]Q=4(-6\exp\left(\frac{-x}{6}\right))\big|^{6}_{0}=-24(\exp(-6/6)-\exp(0))=24(1-\exp(-1))[/tex]
The total charge is Q= 15.171nC.
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 means that the total charge can be calculated as
[tex]Q= \int\limits^6_ 6 2 exp(-|x|/6)dx[/tex]
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
Thus, the total charge is given by:
[tex]Q= 2\int\limits^6_0 2exp (-x/6), dx[/tex]
When we integrate, we get:
Q= 24(1- exp(-1))nC ≈
15.171nC
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