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in Electrostatics by (15 points)
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A charge q is uniformly distributed along an insulating straight wire of length \( 2 l \) as shown in Figure. Find an expression for the electric potential at a point located a distance \( d \) from the distribution along its perpendicular bisector.

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Electric potential due to a small element dx of rod

\(dp = \frac 1{4\pi\epsilon_0} \frac{q\,dx}{2l\sqrt{d^2 + x^2}}\)

total \(pE = \int\limits_{-1}^1 \frac1{4\pi\epsilon_0} \frac{q\,dx}{2l\sqrt{d^2 + x^2}}\)

taking \(cos\theta = \frac d{\sqrt{d^2 + x^2}}\)

\(dx=\frac{d}{sec^2\theta }d\theta\)

and \(x=- 1 \) ⇒ \(\theta = \theta _1 = tan^{-1} (\frac{-1}d)\)

and \(x= 1 \) ⇒ \(\theta = \theta _2 = tan^{-1} (\frac{1}d)\)

\(pE = \int\limits_{\theta_1}^{\theta_2} \frac 1{4\pi\epsilon_0} \frac q{2l} \frac{cos\theta }{d} \frac{d}{cos^2\theta} d\theta\)

\(= \int\limits_{\theta_1}^{\theta_2} \frac 1{4\pi\epsilon_0} \frac q{2l} \,sec\theta \,d\theta \)

\(= \frac1{4\pi\epsilon_0} \frac q{2l} ln\left(sec\theta + tan\theta \right)_{\theta_1}^{\theta_2}\)

\(=\frac1{4\pi\epsilon_0} \frac q{2l} ln\left(\frac{\sqrt{d^2 + x^2}+1}{\sqrt{d^2 + x^2}-1}\right)\)

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