How to use the gradient to find Electric field

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SUMMARY

The discussion focuses on calculating the electric field at the center of a semi-circular rod with a uniform charge distribution. The electric potential at the center is expressed as V = kQ/r or V = k(pi)(lambda), where lambda represents the charge density. The correct electric field expression derived is E = 2k(lambda)/r. The confusion arises from incorrectly applying the gradient of the electric potential, leading to an erroneous result of kQ/r^2. Properly accounting for the gradient along the radius and summing contributions from the semi-circular rod is essential for accurate calculations.

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  • Understanding of electric potential and electric field concepts
  • Familiarity with calculus, specifically gradient operations
  • Knowledge of charge distribution and linear charge density
  • Basic principles of electrostatics and field calculations
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  • Study the derivation of electric fields from electric potentials in electrostatics
  • Learn about the application of gradient in vector calculus
  • Explore the concept of charge density and its implications in electric field calculations
  • Investigate the superposition principle in electric fields for complex charge distributions
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1. A rod carrying a uniform charge distribution is bent into a semi circle with the center on the orgin and a radius R. calculate the Electric field at the center of the semi circle using the electric potential expression found in part a



2. E = -(gradient)V



3. The electric potential at the center is V = kQ/r or k (pi) (lambda) where lambda is the charge density. The correct expression for the electric field at the center of the circle is 2k(lambda)/r. However, I'm finding that simply taking the gradient of the electric potential function gives kQ/r^2 or k(lambda)(pi)/r, which is not correct. I'm afraid I'm making a fundamental mistake with the definition of the gradient and how to account for the changing electric field direction at the center of the semi circle.
 
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Here you have taken the gradient along the radius.
First you have to finds out delta Ex( due to a small element dl) = -dV/dx and dEy = - dv/dy where dx = dr*cos (theta) and dy = dr*sin(theta). Take summation over the whole semicircular rod to find the field.
 

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