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Positive electric charge Q is distributed uniformly along a thin rod of length 2a. The rod lies along the x-axis between x = -a and x = +a (fig. 23.27). Calculate how much work you must do to bring a positive point charge q from infinity to the point x = +L on the x-axis, where:L> a.

Respuesta :

Answer:

kQq(ln(L+2a/L+a))/2a

Explanation:

This question is a example of potential difference calculation.

firstly we have to calculate the potential at point +L which we can find by integration.

steps for finding it

Firstly take a small section of dx length on road of charge (Qdx/2a) which is at x distance from the point +L.

Then find dv due to that using v=kq1/r formula

So required function will be dv=kQdx/(2a*x) we will integrate on both side with limit x=L+2a to x=L+a.

Then multiply v by charge which we will be carrying from infinity to that point.

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