How Does the Earthed Sphere Acquire Charges?

In summary, the distance between the centers of the spheres affects the calculations for the potential at the surfaces of the spheres. When considering the potential at the surface of sphere B and C, the distance should be d-a instead of d+2a. This correction will result in only one possible answer that is close to the given options.
  • #1
Jahnavi
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Homework Statement


spheres.jpg


Homework Equations

The Attempt at a Solution



When a sphere of charge Q is touched to A and removed , A acquires charge Q/2 .

When B is earthed , potential of B becomes zero . B acquires some charge q .

V( B's center ) = 0

kQ/[2(d+2a)] + kq/a = 0

q = -aQ/[2(d+2a)]

Now earthed connection of B is removed and C is earthed .C acquired some charge q' .

V( C's center ) = 0

kQ/[2(d+2a)] -kaQ/[2(d+2a)]2+ kq'/a = 0

q' = -aQ(d+a)/[2(d+2a)]2 . Now I think I need to make use of the approximation a/d<<1 .But this doesn't simplify in any of the options .

Where am I going wrong ?
 

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  • #2
Jahnavi said:

Homework Statement


View attachment 221887

Homework Equations

The Attempt at a Solution



When a sphere of charge Q is touched to A and removed , A acquires charge Q/2 .

When B is earthed , potential of B becomes zero . B acquires some charge q .

V( B's center ) = 0

kQ/[2(d+2a)] + kq/a = 0

q = -aQ/[2(d+2a)]

Now earthed connection of B is removed and C is earthed .C acquired some charge q' .

V( C's center ) = 0

kQ/[2(d+2a)] -kaQ/[2(d+2a)]2+ kq'/a = 0

q' = -aQ(d+a)/[2(d+2a)]2 . Now I think I need to make use of the approximation a/d<<1 .But this doesn't simplify in any of the options .

Where am I going wrong ?
##d \gt\gt a## implies that you can treat the charge on sphere A as a point charge when considering the potential at B due to A. You could make a similar assumption for sphere C.
 
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  • #3
tnich said:
##d \gt\gt a## implies that you can treat the charge on sphere A as a point charge when considering the potential at B due to A. You could make a similar assumption for sphere C.

Do you get one of the options ?
 
  • #4
Jahnavi said:
Do you get one of the options ?
Yes. Just assume the spheres are all point charges relative to each other.
 
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  • #5
tnich said:
Yes. Just assume the spheres are all point charges relative to each other.

Even if I use exact distances , then too , I should get one of the options with appropriate approximation in the expression I derived in the OP . Do you think the expression for charge on C in the OP is incorrect ?
 
  • #6
Jahnavi said:
Even if I use exact distances , then too , I should get one of the options with appropriate approximation in the expression I derived in the OP . Do you think the expression for charge on C in the OP is incorrect ?
I question that the effective distance between the charges on a pair of spheres is ##d+2a##. What is your rationale for that assumption?
 
  • #7
tnich said:
I question that the effective distance between the charges on a pair of spheres is ##d+2a##. What is your rationale for that assumption?

While calculating potential due to a charged sphere at earthed sphere's center , distance will be (d+2a) . We could have also chosen the sphere's surface instead of its center . In that case distance will be (d+a) .
 
  • #8
Jahnavi said:
While calculating potential due to a charged sphere at earthed sphere's center , distance will be (d+2a) . We could have also chosen the sphere's surface instead of its center . In that case distance will be (d+a) .
I see. What if the centers of the spheres are located at the vertices of the triangle?
 
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  • #9
The potential at the surface of ## B ## , and then again, the potential at the surface of ## C ## needs to be zero. The electric field inside of ## B ## and ## C ## will be zero, but you need to establish the zero potential at the surface of each. I think the distance is then ## d-a ## from the center of one sphere to the surface of the other. ## \\ ## Meanwhile, in making the potential at the surface of ## C ## equal to zero in the OP, the "2" in the denominator of the ## Q_B## term belongs outside of the parenthesis, and should not be squared. Once you make that correction, there is only one answer that comes close. (At first, I didn't get their answer exactly). If you set the distance ## d-a ## in the denominators equal to ## d ## instead of ## d-a ## in both cases, then the result is exactly their answer.
 
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  • #10
Charles Link said:
The potential at the surface of ## B ## , and then again, the potential at the surface of ## C ## needs to be zero. The electric field inside of ## B ## and ## C ## will be zero, but you need to establish the zero potential at the surface of each. I think the distance is then ## d-a ## from the center of one sphere to the surface of the other. ## \\ ## Meanwhile, in making the potential at the surface of ## C ## equal to zero in the OP, the "2" in the denominator of the ## Q_B## term belongs outside of the parenthesis, and should not be squared. Once you make that correction, there is only one answer that comes close. (At first, I didn't get their answer exactly). If you set the distance ## d-a ## in the denominators equal to ## d ## instead of ## d-a ## in both cases, then the result is exactly their answer.
If we treat the spheres as point charges relative to each other and assume the distance between the centers of any two of them is ##d##, then the denominators will be ##d## and ##d^2##. We cannot assume that all of the charge on a sphere will concentrate on the point closest to a sphere with a charge of the opposite sign. Even if we did, that point would change depending on which pair of spheres we consider. We also need to consider the repulsive force of the charge on each sphere, which will tend to distribute it evenly over the surface. When ##d## is large relative to ##a##, the difference in electrostatic force between one side of the sphere vs. the other due to a far-off charge is negligible. So I think we can assume that the charge is distributed evenly over the surface of each sphere and centered at the center of each sphere.
 
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  • #11
tnich said:
If we treat the spheres as point charges relative to each other and assume the distance between the centers of any two of them is ##d##, then the denominators will be ##d## and ##d^2##. We cannot assume that all of the charge on a sphere will concentrate on the point closest to a sphere with a charge of the opposite sign. Even if we did, that point would change depending on which pair of spheres we consider. We also need to consider the repulsive force of the charge on each sphere, which will tend to distribute it evenly over the surface. When ##d## is large relative to ##a##, the potential difference between one side of the sphere vs. the other due to a far-off charge is negligible compared. So I think we can assume that the charge is distributed evenly over the surface of each sphere and centered at the center of each sphere.
@tnich The charge on the two spheres must be such that the total electric potential is zero when you are at the surface of the grounded sphere. I suppose the way you worded it will also work. ## \\ ## Editing: In using ## d-a ##, I happened to pick the point on the surface of the grounded sphere nearest the ungrounded sphere, but that is really unjustified, because one could argue, why not instead pick the point on the surface furthest from the ungrounded sphere? Choosing the distance as ## d ## is perhaps the most reasonable choice. So yes, I agree with you. ## \\ ## In any case, the most significant error in the OP's @Jahnavi calculation is the factor of "2 " in the parenthesis, as mentioned in post 9, along with the next algebraic step, where a ## (2d+...) ## term should have arisen in the numerator upon having a ## [2(d+2a)]^2 ## as a common denominator, instead of just ## (d+a) ##. That would have made the correct answer almost obvious, but to get the exact answer, the ## d+2a ## in the denominator needs to be simply ## d ##.
 
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  • #12
Thanks @tnich .

I do get the correct option 3) using your suggestion .

Charles Link said:
The potential at the surface of ## B ## , and then again, the potential at the surface of ## C ## needs to be zero. The electric field inside of ## B ## and ## C ## will be zero, but you need to establish the zero potential at the surface of each. I think the distance is then ## d-a ## from the center of one sphere to the surface of the other.

Since spheres are equipotential , any point on/in the sphere should work . Center could be a choice .

Charles Link said:
## \\ ## In any case, the most significant error in the OP's @Jahnavi calculation is the factor of "2 " in the parenthesis, as mentioned in post 9, along with the next algebraic step, where a ## (2d+...) ## term should have arisen in the numerator upon having a ## [2(d+2a)]^2 ## as a common denominator, instead of just ## (d+a) ##. That would have made the correct answer almost obvious, but to get the exact answer, the ## d+2a ## in the denominator needs to be simply ## d ##.

It is a typing mistake in the OP . In actual work I did , 2 isn't squared .

From the OP , the simplified expression I get is QC = -aQ(d+a)/[2(d+2a)2].

Could you show , how this is approximately equivalent to option 3) ?
 
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  • #13
Jahnavi said:
Thanks @tnich .

I do get the correct option 2) using your suggestion .
Since spheres are equipotential , any point on/in the sphere should work . Center could be a choice .
It is a typing mistake in the OP . In actual work I did , 2 isn't squared .

From the OP , the simplified expression I get is QC = -aQ(d+a)/[2(d+2a)2].

Could you show , how this is approximately equivalent to option 3) ?
Take the limit as ## d>>a ##, and (3) is the only one with the correct sign of the electric charge that isn't off by a factor of 2. Both (2) and (4) have the wrong sign for the electric charge. Answer (1) has the correct sign for the electric charge but it is off by a factor of 2. ##\\ ## Edit: The correction for this ## d+2a ##, ## d+a ## , or ## d-a ##, vs. ## d ## in the denominator is all about the same size as the correction of the effect of the electrical charge at ## B ##, so that unless you use ## d ## for the distance in the denominator, you won't get their exact answer.
 
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  • #14
The question is a bit dodgy. We are told that d>>a, yet the answers offered do not simplify d-a to d, etc. This means that approximations along the way, such as treating the spheres as point charges, could invalidate the answers. To get a safe match on one of the options would require first and second order terms to be retained from the start.
 
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  • #15
@Jahnavi See also the edited part of my post 13. Also see my post 11, middle paragraph, which corrects my result of post 9.
 
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  • #16
Charles Link said:
Take the limit as ## d>>a ##, and (3) is the only one with the correct sign of the electric charge that isn't off by a factor of 2. Both (2) and (4) have the wrong sign for the electric charge. Answer (1) has the correct sign for the electric charge but it is off by a factor of 2. ##\\ ## Edit: The correction for this ## d+2a ##, ## d+a ## , or ## d-a ##, vs. ## d ## in the denominator is all about the same size as the correction of the effect of the electrical charge at ## B ##, so that unless you use ## d ## for the distance in the denominator, you won't get their exact answer.

Thank you very much :smile:
 
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Related to How Does the Earthed Sphere Acquire Charges?

1. What is the charge on an earthed sphere?

The charge on an earthed sphere is zero. This is because an earthed sphere is connected to the ground, which acts as a neutralizing agent for any excess charge on the sphere.

2. How is the charge on an earthed sphere affected by external charges?

The charge on an earthed sphere is not affected by external charges. This is because the ground acts as a conductor and any excess charge on the sphere will be neutralized by the ground.

3. Can the charge on an earthed sphere be changed?

Yes, the charge on an earthed sphere can be changed by adding or removing charges from the sphere. However, as long as the sphere remains connected to the ground, the charge will be neutralized.

4. What is the purpose of having an earthed sphere?

An earthed sphere is used to protect against electric shock. It acts as a grounding point for any excess charge, preventing it from building up and causing harm to individuals or equipment.

5. How does the charge on an earthed sphere affect its electric potential?

The charge on an earthed sphere has no effect on its electric potential. This is because the potential of a conductor is constant and equal to the potential of the earth, which is considered to be zero.

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