Question

Asked Feb 4, 2020

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Two rods of equal length *L* = 1.87 m form a symmetric cross. The horizontal rod has a charge of *Q* = 300.0 C, and the vertical rod a charge of -*Q* = -300.0 C. The charges on both rods are distributed uniformly along their length. Calculate the potential at the point *P* at a distance *D* = 3.05 m from one end of the horizontal rod.

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Step 1

Given:

the length of rod *(L)* = 1.87 m

charge on horizontal rod *(Q _{1} )* = 300 C

charge on vertical rod *(Q _{2} )* = -300 C

distance between horizontal plate and point P (*D*) = 3.05 m

Step 2

charge density of rod = Q/L

potential at point P due to horizontal rod is,

Step 3

distance between any point on rod from point P is hypotenus,

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