Question
Asked Oct 5, 2019
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I do not understand how the professor came to the conclusion for the Vcenter equation.

O Rutgers, The State University of New Jersey. Department of Physics&Astronormy
Example: Find the electric field and the potential
at the center of a thin square conductor of side L
and total charge Q.
Ecenter 0,
Vcenter 0
L/2
8k1
1
xp
:dx
2.
center=
2
+
r2
2
O Rutgers, The State University of New Jersey. Department of Physics & Astranomy
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O Rutgers, The State University of New Jersey. Department of Physics&Astronormy Example: Find the electric field and the potential at the center of a thin square conductor of side L and total charge Q. Ecenter 0, Vcenter 0 L/2 8k1 1 xp :dx 2. center= 2 + r2 2 O Rutgers, The State University of New Jersey. Department of Physics & Astranomy

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Expert Answer

Step 1

The potential at the center of the square would be given by the sum of individual potentials at the center due to conductor segments of length L/2. Consider the diagram below of one such segment.

Step 2

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L/2 dx x L/2

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Step 3

The potential due an length element dx of the conductor at the center is given as,

Here, dq is the charge in dx length, k is the Coulomb’s constant and r is the line element’s distance from the center.

Integrate above equation to obtain the potential at the cen...

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da bp= AP dV k-

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