Electric Dipole Flux at Center of Sphere: Answer & Explanation

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SUMMARY

The electric flux due to an electric dipole placed at the center of a sphere is definitively zero. This conclusion is derived from Gauss' Law, which states that the electric flux through a closed surface is proportional to the total charge enclosed. Since an electric dipole has no net charge (the positive and negative charges cancel each other), the total charge enclosed by the sphere is zero, resulting in zero electric flux. The mathematical representation of this is given by the equation Φ_E = q/ε₀, where q is the enclosed charge.

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  • Understanding of Gauss' Law in electrostatics
  • Familiarity with electric dipoles and their properties
  • Knowledge of electric field concepts and flux calculations
  • Basic mathematical skills for vector calculus
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  • Learn about electric field lines and their relationship to charge distributions
  • Explore the concept of electric dipoles in more complex configurations
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Students of physics, particularly those studying electromagnetism, educators teaching electrostatics, and anyone seeking to deepen their understanding of electric fields and flux concepts.

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Homework Statement


an electrical dipole is placed at the centre of the sphere,the electric flux due to dipole is?
i know the answer is zero but i don't find my reasons satisfactory pls post reply
(i thought if the flux are normal to the centre the cos90 is zero hence zero)
pls pls reply :)
 
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venkat1989 said:

Homework Statement


an electrical dipole is placed at the centre of the sphere,the electric flux due to dipole is?
i know the answer is zero but i don't find my reasons satisfactory pls post reply
(i thought if the flux are normal to the centre the cos90 is zero hence zero)
pls pls reply :)

Do you know Gauss 's Law? What is the charge enclosed in that sphere?
 
hey the sphere has a charge(some 'Q') but now there is a dipole and it has certain charge...
 
venkat1989 said:
hey the sphere has a charge(some 'Q') but now there is a dipole and it has certain charge...

The sphere is a mathematical construct, not a physical entity. It has no charge of its own.

Before you read on, think what the total charge of an electric dipole is, and how that affects the result you'd expect from Gauss' Law.

The q in Gauss' Law:

\Phi_E=\frac{q}{\epsilon_0}

Is the TOTAL charge enclosed by the Gaussian surface (The charge inside the closed surface!), not the charge on the surface itself.If you were to meticulously calculate the electric flux due to the dipole through the sphere enclosing it, using the definition of the E-field flux as: \Phi_E = \oint \vec E \cdot d\vec A you will find the total flux to be 0.
 
Last edited:
Gauss' law is the right way to approach this problem. One way to "understand" Gauss' law is that the "number" of field lines (i.e., flux) that cross through the boundary is proportional to the charge enclosed. For the dipole, every field line that goes out the plus charge goes back into the negative charge. So the net contribution across the whole surface is zero.
 

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