• Support PF! Buy your school textbooks, materials and every day products Here!

MAgnitude of electric Field

  • Thread starter musicman05
  • Start date
1. Homework Statement
A solid sphere of radius 40.0 cm has a total positive charge of 46.0 µC uniformly distributed throughout its volume. Calculate the magnitude of the electric field:

10.0 cm from the center of the sphere
40.0 cm from the center of the sphere
60.0 cm from the center of the sphere



2. Homework Equations

E=kq/r^2



3. The Attempt at a Solution

8.99e9*46e6/60cm=6.892333333e15
 

Answers and Replies

458
0
Let's call the charge enclosed by a sphere of radius r (with r < R) Q', and the charge of the entire sphere, Q.

The ratio of the charges is thus [tex]\frac{Q'}{Q}=\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3}[/tex]

or:

[tex]Q'=Q\frac{r^3}{R^3}[/tex]

Using this, and the fact that [tex]\int \vec{E}.\vec{da}=\frac{q}{\epsilon _0}[/tex] from Gauss's Law.

You can calculate the electric field at any point within the sphere.
 
Last edited:

Related Threads for: MAgnitude of electric Field

  • Last Post
Replies
2
Views
2K
Replies
3
Views
8K
  • Last Post
Replies
1
Views
773
Replies
6
Views
86K
Replies
3
Views
20K
Replies
6
Views
1K
  • Last Post
Replies
1
Views
5K
Top