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Eletric Field of a induced sphere

Physics Asked by Emanuel Dias on July 6, 2021

Suppose I have a conductor sphere, and a point charge at a distance x away from the center of the sphere.

The sphere will be polarized, but what about it field lines of that sphere? Is zero?

I would say yes, with Gauss law, but if is that the case, the point charge will be at equilibrium instead of exist an attractive force. Im really confused.

3 Answers

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Firstly U need to understand that this is the way the charge is going to be induced Now for the flux part

since the charge distribution in the sphere is symmetrical the number of field lines entering is gonna be equal to the number of field lines leaving the gaussian surface hence resulting in zero flux .

But the charge will experience a force because We know that inside the conductor electric field is zero. So, E Net= E ind + E ext =0 ⇒ Eind =−Eext= − Kq/r^2 i cap

so now F = qE

Hope it helps Have fun ELecrostating ;)

Answered by Paras Bisht on July 6, 2021

For a force to exist it is necessary and sufficient to have an electric field at the charge location. A dipole may have a zero net charge, yet it creates a dipole electric field and affects the other charges (as well as itself is affected by the external field sources).

Answered by Vladimir Kalitvianski on July 6, 2021

Suppose I have a conductor sphere, and a point charge at a distance x away from the center of the sphere.

The sphere has no net charge I assume?

The sphere will be polarized, but what about it field lines of that sphere? Is zero?

The net flux from the sphere will be zero with any increase in one side being associated with a decrease elsewhere. But that doesn't mean that the field is equal to zero at every point on the gaussian surface.

I would say yes, with Gauss law, but if is that the case, the point charge will be at equilibrium instead of exist an attractive force. Im really confused.

A net flux of zero from the sphere does not imply zero force on the charge. As the induced charges are polarized, some are closer to the test charge and those have a greater effect on the forces.

Answered by BowlOfRed on July 6, 2021

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