Electric Potential of Uniformly Charged Sphere

In summary, the conversation discusses the calculation of electric potential at different distances from a nonconductive sphere with a given radius and charge. The equations Gauss' Law and V=E*ds are mentioned, but the latter is incorrect as the RHS is a differential element and the LHS is not. The correct equation for electric potential is V=kqr/R. The person asking for help is getting an answer that is 20 times smaller than the actual value and is asked to show their work more explicitly.
  • #1
ktbobabee@aol
1
0

Homework Statement


A nonconductive sphere has radius R=2.31 cm and uniformly distributed charge q==3.5 fC. Take the electric potential at the sphere's center to be V=0. What is V at the radical distance r=1.45cm and at r=2.31 cm


Homework Equations


Gauss' Law
V=E*ds


The Attempt at a Solution


E=kqr/R^3
then V=E*ds
I keep getting an answer that is 20 times smaller then the actual.


 
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  • #2
Your second equation "V=E*ds" is not correct. The RHS is a differential element, while the LHS is not. Besides, what is "s"?

Please show your work more explicitly. With what you've shown so far, there is no way one can tell you exactly where your error is.
 

Related to Electric Potential of Uniformly Charged Sphere

What is the electric potential of a sphere?

The electric potential of a sphere is the measure of the work done to move a unit test charge from infinity to a point on the surface of the sphere, divided by the magnitude of the test charge. In other words, it is the potential energy per unit charge at a specific point on the surface of the sphere.

How is the electric potential of a sphere calculated?

The electric potential of a sphere can be calculated using the equation V = kQ/r, where V is the electric potential, k is the Coulomb's constant, Q is the charge of the sphere, and r is the distance from the center of the sphere to the point where the potential is being measured.

What factors affect the electric potential of a sphere?

The electric potential of a sphere is affected by the charge of the sphere, the distance from the center of the sphere, and the surrounding medium. The potential also decreases with increasing distance from the surface of the sphere.

Why is the electric potential of a conductor constant on its surface?

According to the properties of conductors, the electric field inside a conductor is zero, meaning that there is no potential difference. Therefore, the potential on the surface of the conductor must be constant and equal to the potential of the surrounding medium. This is known as the "boundary condition" for conductors.

How does the electric potential of a sphere relate to its electric field?

The electric potential of a sphere is directly related to its electric field. The electric field is the negative gradient of the electric potential, meaning that it is the rate of change of potential with respect to distance. Therefore, a higher electric potential will result in a stronger electric field and vice versa.

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