Question #262253

What is the total charge on an infinite wire carring a charge density


λ =


1


|x|+α where α is a cosntant.

1
Expert's answer
2021-11-08T09:12:53-0500

Given,

Charge density of the charged wire (λ)=1x+α(\lambda) = \frac{1}{|x|+\alpha}

dq=λdx\Rightarrow dq=\lambda dx

dq=λdx\Rightarrow \int dq = \int \lambda dx

Now, let the total length of the wire be l. So, integration limit will be l-l to +l+l

q=0q=Qdq=llλdx\Rightarrow \int_{q=0}^{q=Q} dq=\int_{-l}^{l}\lambda dx

Now, taking the integration

Q=x=0x=ldxxαx=0x=ldxx+αQ=\int_{x=0}^{x=l} \frac{dx}{x-\alpha}-\int_{x=0}^{x=l} \frac{dx}{x+\alpha}


Q=[ln(xα)]0lln(x+α)]0l\Rightarrow Q = [\ln(x-\alpha)]_{0}^{l}-\ln{(x+\alpha)}]_{0}^{l}


Q=ln(xα)ln(x+α)\Rightarrow Q = \ln(x-\alpha)-ln(x+\alpha)


Q=ln(xα)(x+α)\Rightarrow Q = \ln{\frac{(x-\alpha)}{(x+\alpha)}}


Q=ln(x(1αx)x(1+αx))\Rightarrow Q = \ln(\frac {x(1-\frac{\alpha}{x})}{x(1+\frac{\alpha}{x})})

xx\rightarrow \infty

Q=ln10\Rightarrow Q = \ln 1\sim 0

Hence, net charge on the straight infinite wire be zero.


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