Potential difference between the center of a sphere and a point 4.0 cm away?

In summary, the conversation discusses the use of integral to calculate the potential (V) of a nonconducting sphere with a charge Q, and the confusion surrounding the use of the equation V = KQ/r to obtain the correct answer. It is explained that the equation only gives the potential at a specific point, and to find the potential difference between two points, the integral of the electric field (E) must be used. The concept of potential being relative and the potential due to all charges present is also mentioned.
  • #1
hidemi
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Homework Statement
A nonconducting sphere of radius 10 cm is charged uniformly with a density of 100 nC/m^3. What is the magnitude of the potential difference between the center and a point 4.0 cm away?

The answer is 3.0 V.
Relevant Equations
V = kQ / r
I can use integral as attached to obtain the answer.
However, I wonder why I can not use " V = KQ / r " the get the correct answer.
Thanks.
 

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  • #2
hidemi said:
However, I wonder why I can not use " V = KQ / r " the get the correct answer.
Thanks.
How exactly did you use that? E.g. what value for Q?
 
  • #3
The Q value I plug in is density*4pi/3*r^3, so V = k (density*4pi/3*r^2)
My confusion is that instead of using V = kQ/r directly, we integrate electric field (E) to calculate the potential (V) to obtain the correct answer?
 
  • #4
hidemi said:
The Q value I plug in is density*4pi/3*r^3, so V = k (density*4pi/3*r^2)
My confusion is that instead of using V = kQ/r directly, we integrate electric field (E) to calculate the potential (V) to obtain the correct answer?
kQ/r would be the potential at a point P at radius r, ignoring any charges outside radius r, and taking the potential at infinity to be zero. This is not the same as finding the potential difference between P and the origin, O.
There is a potential due to the charges beyond radius r, but that is the same for both P and O. The difference is that the sphere within radius r creates a potential at O. Since you want the difference in potentials, this needs to be subtracted from kQ/r.
 
  • #5
haruspex said:
kQ/r would be the potential at a point P at radius r, ignoring any charges outside radius r, and taking the potential at infinity to be zero. This is not the same as finding the potential difference between P and the origin, O.
There is a potential due to the charges beyond radius r, but that is the same for both P and O. The difference is that the sphere within radius r creates a potential at O. Since you want the difference in potentials, this needs to be subtracted from kQ/r.
I know V at origin is 0 for nonconducting sphere. I just wonder why can't we plug in 0.04m directly into the equation to get the potential difference, instead we need to integrate the electric field from 0-0.04m to calculate the potential?
 
  • #6
$$V(B)-V(A)=-\int_{\vec r_A}^{\vec r_B} \vec E \cdot d\vec l$$
Let's compute ##\vec E##
 
  • #7
hidemi said:
I know V at origin is 0 for nonconducting sphere.
No, the field there is zero.
Potential is, in principle, relative. A common standard is to take the potential at infinity to be zero. Everywhere else, it is the sum of potentials due to all charges present.
If the only charge being considered is Q on a spherical shell radius R then the potential everywhere in the shell is kQ/R.
 
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  • #8
haruspex said:
No, the field there is zero.
Potential is, in principle, relative. A common standard is to take the potential at infinity to be zero. Everywhere else, it is the sum of potentials due to all charges present.
If the only charge being considered is Q on a spherical shell radius R then the potential everywhere in the shell is kQ/R.
Thanks a lot.
 

Related to Potential difference between the center of a sphere and a point 4.0 cm away?

1. What is potential difference?

Potential difference is the difference in electric potential between two points in an electric field. It is a measure of the work required to move a unit of electric charge from one point to another.

2. How is potential difference calculated?

Potential difference is calculated by dividing the work done in moving a unit of charge from one point to another by the magnitude of the charge. In mathematical terms, it is written as V = W/Q, where V is the potential difference, W is the work done, and Q is the charge.

3. What is the potential difference between the center of a sphere and a point 4.0 cm away?

The potential difference between the center of a sphere and a point 4.0 cm away depends on the charge of the sphere and the distance between the two points. It can be calculated using the formula V = kQ/r, where k is the Coulomb's constant, Q is the charge of the sphere, and r is the distance between the two points.

4. How does the potential difference change as the distance from the center of the sphere increases?

As the distance from the center of the sphere increases, the potential difference decreases. This is because the electric field strength decreases with distance, leading to a decrease in the work required to move a unit of charge from the center to the point 4.0 cm away.

5. What factors affect the potential difference between the center of a sphere and a point 4.0 cm away?

The potential difference between the center of a sphere and a point 4.0 cm away is affected by the charge of the sphere, the distance between the two points, and the medium between them. It is also influenced by any other charges or conductors nearby, as they can alter the electric field and affect the potential difference.

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