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The rod is of length L and has uniform charge per unit length 2. Find the electric potential at point P on the y-axis. Use dq = Adx r= Vx2 + d2 uleyods no Adx V = ke | Vx2 + d² d Since this is not an elementary integral, you will have to consult an integral table.

The rod is of length L and has uniform charge per unit length 2. Find the electric potential at point P on the y-axis. Use dq = Adx r= Vx2 + d2 uleyods no Adx V = ke | Vx2 + d² d Since this is not an elementary integral, you will have to consult an integral table.

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**Question 4: Potential for a Finite Line of Uniform Charge at P** The rod is of length \( l \) and has uniform charge per unit length, \(\lambda\). Find the electric potential at point \( P \) on the y-axis. Use \( dq = \lambda dx \) \( r = \sqrt{x^2 + d^2} \) \[ V = k_e \int_{0}^{l} \frac{\lambda dx}{\sqrt{x^2 + d^2}} \] Since this is not an elementary integral, you will have to consult an integral table. --- **Diagram Explanation:** The diagram illustrates a rod of length \( l \) positioned horizontally, with point \( P \) located at a perpendicular distance \( d \) from the rod, on the y-axis. The line connecting a point \( x \) on the rod to point \( P \) represents the distance \( r = \sqrt{x^2 + d^2} \).
Transcribed Image Text:**Question 4: Potential for a Finite Line of Uniform Charge at P** The rod is of length \( l \) and has uniform charge per unit length, \(\lambda\). Find the electric potential at point \( P \) on the y-axis. Use \( dq = \lambda dx \) \( r = \sqrt{x^2 + d^2} \) \[ V = k_e \int_{0}^{l} \frac{\lambda dx}{\sqrt{x^2 + d^2}} \] Since this is not an elementary integral, you will have to consult an integral table. --- **Diagram Explanation:** The diagram illustrates a rod of length \( l \) positioned horizontally, with point \( P \) located at a perpendicular distance \( d \) from the rod, on the y-axis. The line connecting a point \( x \) on the rod to point \( P \) represents the distance \( r = \sqrt{x^2 + d^2} \).
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