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Buffu

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## Homework Statement

A line charge with uniform charge distribution and linear charge density ##\lambda## lie along z-axis from ##(0,0,-d)## to ##(0,0,+d)##. Find electric field due to this charge along z, x-axis and potential at ##(0,0,2d)##, ##(\sqrt{3}d, 0,0)## and ##(3d/2, 0 , d)##.

## Homework Equations

## The Attempt at a Solution

##\displaystyle dE_z = \dfrac{dQ}{R^2} = {\lambda dR \over R^2}##

Hence, ##\displaystyle E_z = \int^{z+d}_{z-d} {\lambda dR \over R^2} = {2\lambda d \over z^2 - d^2}##

Where ##z## is a point along z-axis.Now for potential

##\displaystyle \phi_z = -\int^z_0 E_z dz = -\int^z_0 {2\lambda d \over z^2 - d^2} dz = -\lambda\ln \left| d^2 (z-d) \over z + d\right|##

Now for x-axis,

##\displaystyle dE_x = {dQ \cos \theta \over R^2} = {\lambda dz \cos \theta \over R^2} = {\lambda dz \cos \theta \over x^2 + z^2} = {\lambda x \over (x^2 + z^2)^{3/2}}dz##

So, ##\displaystyle E_x = \int^{d}_{-d}{\lambda x \over (x^2 + z^2)^{3/2}} dz = {2\lambda d\over x(d^2 + x^2)^{1/2}}##

Now for potential,

##\displaystyle \phi_x = -\int^x_0 {2\lambda d \over x(x^2 + d^2)^{1/2}} = -\lambda \ln\left| \sqrt{d^2 +x^2} - d \over \sqrt{x^2 + d^2 } + d\right|##

Now we need to find potential at ##(0,0,2d)##, ##(\sqrt{3}d, 0,0)## and ##(3d/2, 0 , d)##,

Plugging ##(0,0,2d)## in ##\phi_z## I got ##\lambda \ln 3##.

Plugging ##(\sqrt{3} \times d,0,0)## in ##\phi_x## I got ##\lambda \ln 3##.

So far so good,

For potential at ##(3d/2, 0 , d)##, I plugged this coordinate in

##\displaystyle \phi (x,0,z) =\phi_x + \phi_z = -\lambda \ln\left| \sqrt{d^2 +x^2} - d \over \sqrt{x^2 + d^2 } + d\right| + -\lambda\ln \left| d^2 (z-d) \over z + d\right|##

Now as you see for ##z = d## the expression on the far right is undefined because ##\ln 0## is undefined.

What to do now ? I got every answer correct except this one. The given answer is ##\lambda\ln 3##.

What did I do wrong ?

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