Induced Charges on Uncharged Conductors

[Vibhor]

Homework Statement

Q If an object having charge 'q' is brought near an uncharged conductor (sphere) , would the charge induced on the nearby surface ( closer to the charged object ) be equal to '-q' ?

Homework Equations

The Attempt at a Solution

I think the charge induced would be less than '-q' . There is no reason for equal magnitude charge ( -q ) to be induced on the nearby surface of the neutral conductor .

Is that correct ?

Thanks

 

[ehild]

What do you mean on the "nearby surface "? If a charged object is brought near to an uncharged one, the charge is redistributed on both (conductive) objects. Assuming a point charge, the new charge distribution of the uncharged conductor is so that the surface of the conductor is equipotential.
In cases the uncharged object is grounded, it gains charge from the ground.

 

[Vibhor]

Assuming a point charge, the new charge distribution of the uncharged conductor is so that the surface of the conductor is equipotential.

Assuming a point charge 'q' near a conducting sphere , would the charge induced on the surface of sphere closer to the point charge be '-q' ?

 

[ehild]

Assuming a point charge 'q' near a conducting sphere , would the charge induced on the surface of sphere closer to the point charge be '-q' ?

I do not know. You can calculate the electric field by using the mirror charge method, placing a virtual negative charge inside the sphere, and a point charge equal to q in the centre of the sphere. Calculate the resultant electric field on the surface of the sphere. The induced charge density at a point is equal Eε₀. If you integrate the surface charge density over the surface closer to the point charge than the other part of the sphere is, you get the answer. I never tried. Certainly, the points close to the point charge become negative, and the points far from the point charge become positive, but I do not know where the border line is.

 

[haruspex]

Assuming a point charge 'q' near a conducting sphere , would the charge induced on the surface of sphere closer to the point charge be '-q' ?

More or less, yes. If it is very close then we can treat the sphere as an infinite plane. By the method of images, the plane conductor at distance x from q is equivalent to a point charge -q at distance 2x. The lines of flux from the point charge all terminate at the plane, implying a total charge of -q there. But exactly how it is distributed I do not know.

 

[ehild]

The case is simple when a plate is involved. The front surface (closer to the point charge) becomes negative, the back surface (farther from the point charge) becomes positive. You get an inhomogeneous charge distribution on the front surface and a homogeneous one on the back surface.

 

[haruspex]

The case is simple when a plate is involved. The front surface (closer to the point charge) becomes negative, the back surface (farther from the point charge) becomes positive. You get an inhomogeneous charge distribution on the front surface and a homogeneous one on the back surface.

Surely if it is an infinite plate the positive charges run off to infinity. Certainly for a flat plate approximation to a large sphere there would no charge of the far surface. For a finite plate, I still doubt there would be much in the way of charge on the reverse of the plate near the point charge.
You wrote that the induced charge density is proportional to the inducing field. How does that work when the surface is not normal to the field? Is it proportional to the normal component of the field? That gives a sensible-looking answer for a plate. If the point charge is at distance a from the plate's nominal centre then at radius r from that centre we would have density $\frac{qa}{2\pi(a^2+r^2)^{3/2}}$.

 

[ehild]

You wrote that the induced charge density is proportional to the inducing field. How does that work when the surface is not normal to the field? Is it proportional to the normal component of the field? .

I meant that the induced charge density is proportional to the resultant field at the surface. Being a conductor, it is equipotential and the field is perpendicular to the surface.

 

[haruspex]

I meant that the induced charge density is proportional to the resultant field at the surface. Being a conductor, it is equipotential and the field is perpendicular to the surface.

I can see that the resultant field is perpendicular to the surface, but it's not so clear that the charge density is proportional to the resultant field. I would have expected it to be proportional to the induced field, i.e. the field due to the induced charges.

 

[ehild]

The metal plate was neutral originally, so the surface charge appearing on it due to the point charge is induced charge.
You can determine the electric field of the plate and the point charge near it, by the method of mirror charges, for example. From the electric field, you get the surface charge density as σ=ε₀E.