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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. help_outlineImage Transcriptionclose+ + + + + + + + + + + + -D P. fullscreen
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Step 1

Given:

the length of rod (L) = 1.87 m

charge on horizontal rod (Q1 ) = 300 C

charge on vertical rod (Q2 ) = -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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