Electric Dipole Flux at Center of Sphere: Answer & Explanation

  • Thread starter Thread starter venkat1989
  • Start date Start date
  • Tags Tags
    Dipole Flux
AI Thread Summary
The electric flux due to an electric dipole placed at the center of a sphere is zero, as confirmed by Gauss's Law. This is because the total charge enclosed by the sphere is zero, given that a dipole consists of equal and opposite charges. The flux calculation shows that field lines emanating from the positive charge return to the negative charge, resulting in no net flux through the surface. The reasoning aligns with the definition of electric flux, which considers the total charge enclosed. Thus, the conclusion that the electric flux is zero is accurate and supported by the principles of electrostatics.
venkat1989
Messages
2
Reaction score
0

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 :)
 
Physics news on Phys.org
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top