Calculate the y component of the electric field at a generic point(x,y)

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SUMMARY

The discussion focuses on calculating the y component of the electric field generated by an electric dipole positioned in the x-y plane, with charges +q at (+a, 0) and -q at (-a, 0). The electric field E is derived from the gradient of the electric potential V, expressed as E = -gradV = q/(4πε₀)(1/r²) r-hat. The solution suggests using the electric field equation from point charges rather than the potential to simplify calculations, leading to E = q/(4πε₀) [(1/(r₊² + (x+a)²)) + (1/(r₋² + (x-a)²))]. The user expresses confusion about vector addition in this context.

PREREQUISITES
  • Understanding of electric dipoles and their configuration
  • Familiarity with vector calculus, specifically gradient operations
  • Knowledge of electric field equations for point charges
  • Basic principles of electrostatics, including Coulomb's law
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Homework Statement



an electric dipole is placed in the x-y plane, with the positive charge +q placed in position (+a, 0) and the negative charge -q placed in position (-a, 0)

define r(subscript +) and r(subscript -) the vectors connecting the point defined by r and the two points (+a, 0) and (-a, 0), and r(subscript+) and r(subscript -) their magnitudes; calculate the y component of the electric field at a generic point (x, y) of the plane as a function of r(subscript +) and r(subscript -) (it is quicker to use directly the field equation from a point charge than using the potential)

The Attempt at a Solution



E=-gradV=q/(4 pi epsilon0)(1/(r^2)) r-hat

E=q/(4pi epsilon0) [(1/([r(+)]^2 +(x+a)^2))+(1/([r(-)]^2 + (x-a)^2))]
 
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In the above was a hint:

"... (it is quicker to use directly the field equation from a point charge than using the potential) ... "

The first line of your solution you have E=-gradV ...

I think you are to just add two electric vectors?
 
sorry, I still don't understand. where have I gone wrong?
 

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