Calculating the Electric Potential from an Electric Field problem

In summary, the question asks for the electric potential at two different radial distances from a nonconducting sphere with a radius of 2.31cm and a charge of +3.50 fC. The electric potential at the sphere's center is taken to be 0. Using the given equations, the potential is calculated to be -5.37 x 10^-4 V at a radial distance of 1.45cm and -0.001V at a distance of 2.31cm. However, there seems to be confusion about the value of ds in the integration formula, with the student manual using r/2 as the value while the student initially thought it was equal to r. After dividing the answers by 2
  • #1
frankfjf
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0

Homework Statement



A nonconducting sphere has radius R = 2.31cm and uniformly distributed charge q = +3.50 fC. Take the electric potential at the sphere's center to be Vo = 0. What is V at radial distance (a) r = 1.45cm and (b) r = R?

Homework Equations



V = the negative integral from i to f of E * ds

For a, E is kqr/R^3.

For b, E is kq/r^2

The Attempt at a Solution



When I plug all this in I get -5.37 x 10^-4 V and -.001V, respectively. However, when I look at the answer explanation in the student manual, it seems to indicate that ds (or in this case rs) = r/2. Why is it r/2? How is the book getting that for ds? At first I thought ds was equal to r, but I guess I'm not grasping something. Could someone explain why this is so? I basically have the right answer, but only if I divide both my answers by 2.
 
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  • #2
what are your limits of integration and what's your formula for V after you integrate?
 
  • #3




Hello,

Thank you for reaching out for clarification on this problem. It seems that the book is using a different approach to calculate the electric potential in this problem. Let me explain how I would approach it, and hopefully that will help you understand the discrepancy.

To calculate the electric potential at a point, we use the equation V = -∫E * ds, where E is the electric field and ds is the infinitesimal displacement along the path from the initial point to the final point. In this case, we can use this equation to calculate the electric potential at points (a) and (b) by considering the electric field at those points.

For part (a), the electric field at a distance r = 1.45 cm from the center of the sphere can be calculated using the equation E = kq/r^2, where k is the Coulomb's constant, q is the charge on the sphere, and r is the distance from the center. Plugging in the values given in the problem, we get E = 6.41 x 10^5 N/C. Now, to calculate the electric potential at this point, we need to integrate this electric field over the distance ds, which is equal to r in this case. So, the electric potential at point (a) would be V = -∫E * ds = -∫(6.41 x 10^5 N/C) * (1.45 cm) = -9.29 x 10^5 V. This is the correct answer and does not require dividing by 2.

For part (b), we can use the same approach. The electric field at the surface of the sphere (r = R = 2.31 cm) can be calculated using the equation E = kq/R^3, giving us E = 1.02 x 10^6 N/C. Now, to calculate the electric potential at this point, we integrate over the distance ds, which is equal to R in this case. So, the electric potential at point (b) would be V = -∫E * ds = -∫(1.02 x 10^6 N/C) * (2.31 cm) = -2.36 x 10^6 V. Again, this is the correct answer and does not require dividing by 2.

I'm not sure why the book is using r/2 for the distance ds
 

1. How do you calculate the electric potential from an electric field problem?

To calculate the electric potential from an electric field problem, you can use the formula V = -∫E•dl, where V is the electric potential, E is the electric field, and dl is a small displacement along the path of integration. This integral is also known as the line integral of the electric field.

2. What is the unit of electric potential?

The unit of electric potential is volts (V) or joules per coulomb (J/C). This represents the amount of work needed to move a unit of positive charge from one point to another in an electric field.

3. Can you have a negative electric potential?

Yes, electric potential can be positive, negative, or zero. A negative electric potential indicates that the electric field is doing work on the charge as it moves from one point to another, meaning the charge is losing potential energy. A positive electric potential means the charge is gaining potential energy.

4. How does distance affect electric potential?

The electric potential is directly proportional to the distance from the source of the electric field. This means that as the distance increases, the electric potential decreases. This relationship is described by the inverse square law, where the electric potential is inversely proportional to the square of the distance.

5. Can you calculate the electric potential at a specific point in an electric field?

Yes, you can calculate the electric potential at a specific point in an electric field by using the formula V = -∫E•dl and plugging in the values for the electric field and the path of integration. This will give you the electric potential at that point, which represents the potential energy per unit charge at that location.

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