Electric potential at surface sphere problem?

[PhysicStud01]

Homework Statement

Homework Equations

The sphere is positively charged and appears to have its charge concentrated at its centre.
Potential V is calculate to be 285V

The Attempt at a Solution

The charge induced on inside of box is negative.
THe potential is less. It is said that it's being the system is not isolated.

I can't understand these.
"The charge induced on inside of box is negative."
"THe potential is less. It is said that it's being the system is not isolated."

they are the answers given to the last part. can you explain it clearly to me please

 

[gneill]

I can't understand these.
"The charge induced on inside of box is negative."

Can you think of a reason why negative charges might be attracted to the inside surface of the box?
Where might they come from?

"THe potential is less. It is said that it's being the system is not isolated."

Can you identify any places in the diagram where parts of the "system" are open to indefinite influences? Hint: what happens to electric field lines that don't terminate inside the system?

 

[PhysicStud01]

Can you think of a reason why negative charges might be attracted to the inside surface of the box?
Where might they come from?

negative charges are from electrons in the metal box. they just move on the inside of the box. they are attracted be\cause of the sphere.

Can you identify any places in the diagram where parts of the "system" are open to indefinite influences? Hint: what happens to electric field lines that don't terminate inside the system?

these are the top 3 lines that do not touch the metal box. they just go to infinity.


but still i can't understand the reasons given for the answers.
could you please explain to me the concept of electric potential in a clear way. not just the definition usually given by teachers

 

[gneill]

negative charges are from electrons in the metal box. they just move on the inside of the box. they are attracted be\cause of the sphere.

Ah, but the box itself is not isolated. Did you notice the ground connection? What effect might that have on things?

Do all the negative charges that migrate to the inside surface of the box come from the box itself?

these are the top 3 lines that do not touch the metal box. they just go to infinity.

True. They either go to infinity or terminate on some charge elsewhere that is not part of the system under consideration. But the main reason why the system is not isolated is the ground connection. Ground represents a fixed zero potential reference and is an "infinite well" of charges of either sign that can be sourced or sunk there.

Suppose that in the beginning you had the charged sphere separated form the box by a large distance, and the box was isolated from ground and was neutral (no net positive or negative charge, equal numbers of protons and electrons in the box). If you then lowered the sphere into the box the charge on the sphere would induce a charge in the box, separating some of its positive and negative charges so that some electrons would migrate to the inner surface, being attracted by the sphere's positive charge, and leaving behind some unpaired positive charges that migrate to the outer surface of the box. Note the total number of charges in the system remains the same; there has just been some separation and rearrangement of charges due to the influence of the sphere's charge.

Now can you describe what would happen if the box is then connected to ground as depicted in the diagram? Will there be more or less charges on the box? What sign and where will they be located?

could you please explain to me the concept of electric potential in a clear way. not just the definition usually given by teachers

I don't know if I can do better than the actual definition. But I can give you a few hints. Electrical potential energy is a similar concept to gravitational potential energy: It represents the potential energy that a charge has simply due to its particular location in a field. The value of that potential energy can determined by evaluating the amount of work required to move a small positive test charge from infinity to that location under the influence of the field. So the units of electrical potential is the work (energy in Joules) per Coulomb of the test charge, that is J/C. We call this particular combination of units a Volt. So one volt equals one joule per coulomb (V = J/C).

Every charge creates an electric field, and the potential at a given location is the sum of the contributions of every charge's field. Each charge's contribution at a given location depends upon the sign of the charge, its magnitude, and the distance from the charge (So there's a big hint here for the given problem --- how many charges are involved in determining the net potential at the surface of the sphere, and where are they located?).

 

[PhysicStud01]

Ah, but the box itself is not isolated. Did you notice the ground connection? What effect might that have on things?

i think this will cause the outside of the box to be neutral while the inside is negatively charged.

But the main reason why the system is not isolated is the ground connection.

is the reason the system is not isolated the ground or the box? what if as you said before, the box was not grounded, would that mean that the sphere is isolated? does the box not affect it?

Now can you describe what would happen if the box is then connected to ground as depicted in the diagram? Will there be more or less charges on the box? What sign and where will they be located?

there would be more charges becuase the inside is negatively charged causing the outside to be positively charged. Now, when grounded, negative charges are added to the outside, causing it to become neutral. so there is more charges overall on the box

how many charges are involved in determining the net potential at the surface of the sphere, and where are they located

it's all the charges on the spheres only, i think. or does the charges of the inside of the box also contribute?

Each charge's contribution at a given location depends upon the sign of the charge, its magnitude, and the distance from the charge (So there's a big hint here for the given problem

SOrry, I tried reading this line several times, but still can't come up with the trick



also, i read somewhere else that in this case, the sphere which is positively charged would tend to repel the unit positive charge, while the negative inside of the box would tend to attract it. this makes the energy required to bring the unit positive charge from infinity to the surface of the sphere smaller (energy required is smaller). So, electric potential is smaller.
Is this correct?

 

[gneill]

i think this will cause the outside of the box to be neutral while the inside is negatively charged.

Correct.

is the reason the system is not isolated the ground or the box? what if as you said before, the box was not grounded, would that mean that the sphere is isolated? does the box not affect it?

It is the ground that makes the system not isolated. The sphere and box taken together, alone without a ground connection, would be isolated: no way for charge to enter or leave the system, making the total charge constant. With the ground connection, the total charge on the box can change.

there would be more charges becuase the inside is negatively charged causing the outside to be positively charged. Now, when grounded, negative charges are added to the outside, causing it to become neutral. so there is more charges overall on the box

Correct.

it's all the charges on the spheres only, i think. or does the charges of the inside of the box also contribute?

Electric potential is the result of the contributions of ALL charges. If the box was isolated then the net charge on the box would be constant, and the only contribution of its charges to the potential near the sphere would be due to the small distance difference of the separated charges on the box's surfaces (the negative charges on the inner surface being slightly closer to the sphere than the positive charges on the box's outer surface). But when the ground is connected, the positive charges on the box are quickly neutralized, leaving the net negative charge alone. The positive charges will no longer partially cancel the contributions of the negative charges.

SOrry, I tried reading this line several times, but still can't come up with the trick

It's pretty much explained in my paragraph above. Bigger charges make a bigger contribution, closer charges make a bigger contribution, and the sign of the charge determines the sign of the contribution. The contribution of paired(net neutral) charges cancel each other. If you separate hem then the closer one's contribution is greater. The induced separated charges on the isolated box will have many negative charges closer to the sphere than positive charges, so the net contribution of potentials from the box's charges will tend to lower the net potential near the sphere a bit.

Grounding the box removes the positive charge from the box and leaves the negative charge at the inner surface (greater than before, since the attraction of the positive charges at the outer surface goes away leaving the sphere's charge as the only source of attraction; more negative charge will move to the inner surface). The net contribution of the charges on the box to the potential at locations near the sphere will be much greater.

also, i read somewhere else that in this case, the sphere which is positively charged would tend to repel the unit positive charge, while the negative inside of the box would tend to attract it. this makes the energy required to bring the unit positive charge from infinity to the surface of the sphere smaller (energy required is smaller). So, electric potential is smaller.
Is this correct?

Yes, that's quite correct.

 

[PhysicStud01]

Electric potential is the result of the contributions of ALL charges. If the box was isolated then the net charge on the box would be constant, and the only contribution of its charges to the potential near the sphere would be due to the small distance difference of the separated charges on the box's surfaces (the negative charges on the inner surface being slightly closer to the sphere than the positive charges on the box's outer surface). But when the ground is connected, the positive charges on the box are quickly neutralized, leaving the net negative charge alone. The positive charges will no longer partially cancel the contributions of the negative charges.

but when using the formula to calculate the the potential V, the inside of the box which is negatively charged is not considered. is it?

It's pretty much explained in my paragraph above. Bigger charges make a bigger contribution, closer charges make a bigger contribution, and the sign of the charge determines the sign of the contribution. The contribution of paired(net neutral) charges cancel each other. If you separate hem then the closer one's contribution is greater. The induced separated charges on the isolated box will have many negative charges closer to the sphere than positive charges, so the net contribution of potentials from the box's charges will tend to lower the net potential near the sphere a bit.

Grounding the box removes the positive charge from the box and leaves the negative charge at the inner surface (greater than before, since the attraction of the positive charges at the outer surface goes away leaving the sphere's charge as the only source of attraction; more negative charge will move to the inner surface). The net contribution of the charges on the box to the potential at locations near the sphere will be much greater.

let's say that instead of being negatively charged, the inside of the box was positively charged, then potential would have been greater, right?


also, from what i read, a system is said to be isolated whenever the total charge in it is constant, right? if yes, are there others conditions?

 

[gneill]

but when using the formula to calculate the the potential V, the inside of the box which is negatively charged is not considered. is it?

Yes! Potential is associated with a location in space. The potential at a given location is due to the contributions of ALL electric charges.

let's say that instead of being negatively charged, the inside of the box was positively charged, then potential would have been greater, right?

Correct.

also, from what i read, a system is said to be isolated whenever the total charge in it is constant, right? if yes, are there others conditions?

Any external influences that might affect the charges would make the system not isolated. So external electric fields, for example, would influence the charges in the system.

 

[PhysicStud01]

Yes! Potential is associated with a location in space. The potential at a given location is due to the contributions of ALL electric charges.

i mean the formula V = Q / 4(pi)er
charge of sphere and its radius are given and used in the formula. but no information about the charge of the inside of box is included. so how does this also take into account all the contributions.

would the potential obtained from calculations be equal whether or not the box was present?

 

[gneill]

i mean the formula V = Q / 4(pi)er
charge of sphere and its radius are given and used in the formula. but no information about the charge of the inside of box is included. so how does this also take into account all the contributions.

It doesn't. If the charged sphere were all alone in space then that formula would indeed yield the potential at its surface.
Because it isn't alone in this case, the potential will end up being less than that given by the formula due to the contributions of the other charges. You can't say how much less, since you don't have details about the box's dimensions, but you can still say that it will be less. So you at least have a qualitative answer if not a quantitative one.

would the potential obtained from calculations be equal whether or not the box was present?

See above.