Find the electric field at a point outside of an insulating sphere?

Click For Summary
SUMMARY

The electric field at a point 0.12m outside the surface of an isolated plastic sphere with a diameter of 0.23m is calculated using the formula E = kQ/(r^2). The charge Q of the sphere is determined to be 1.98375 x 10^-9 C. When substituting this value into the formula with a distance of 0.12m, the calculated electric field is 1239.8 N/C. However, this calculation is incorrect due to the misinterpretation of the distance from the point charge, which should account for the radius of the sphere.

PREREQUISITES
  • Understanding of electric fields and point charges
  • Familiarity with Coulomb's law and the constant k (9 x 10^9 Nm²/C²)
  • Ability to perform calculations involving exponents and scientific notation
  • Knowledge of the geometry of spheres and distances in electrostatics
NEXT STEPS
  • Review the concept of electric fields generated by point charges
  • Learn about the implications of distance in electric field calculations
  • Study the properties of insulating materials and their effect on charge distribution
  • Explore the application of Gauss's law in calculating electric fields
USEFUL FOR

Students studying electrostatics, physics educators, and anyone involved in solving problems related to electric fields and charge distributions.

hockey
Messages
4
Reaction score
0

Homework Statement


What is the electric field at a point .12m outside the surface of an isolated plastic sphere .23m in diameter that produces an electric field of 1350 N/C just outside the surface of the sphere?


Homework Equations


E = kQ/(r^2)


The Attempt at a Solution


First I found Q for the sphere with the above formula by substitution and found that it was 1.98375*10^-9C. From there, I thought that I could treat the sphere as a point charge and use the same formula to find the electric field, substituting as follows:
E= (9*10^9 Nm^2/C^2) (1.98375*10^-9C)/(.12m)^2 and got a result of 1239.8 N/C. This is not correct however, and I am not sure if it is because I am viewing the question wrong or not.
 
Physics news on Phys.org
hockey said:

Homework Statement


What is the electric field at a point .12m outside the surface of an isolated plastic sphere .23m in diameter that produces an electric field of 1350 N/C just outside the surface of the sphere?


Homework Equations


E = kQ/(r^2)


The Attempt at a Solution


First I found Q for the sphere with the above formula by substitution and found that it was 1.98375*10^-9C. From there, I thought that I could treat the sphere as a point charge and use the same formula to find the electric field, substituting as follows:
E= (9*10^9 Nm^2/C^2) (1.98375*10^-9C)/(.12m)^2 and got a result of 1239.8 N/C. This is not correct however, and I am not sure if it is because I am viewing the question wrong or not.

Yes, you can treat the charged sphere as a point charge. But take a closer look at what you used as the distance from that charge for your second calculation. That distance was specified to be 0.12m outside of the original sphere... so where does that put it with respect to the "new" point charge?
 
  • Like
Likes   Reactions: 1 person

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
23
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
11
Views
3K
Replies
12
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K