# MAgnitude of electric Field

1. Apr 2, 2008

### musicman05

1. The problem statement, all variables and given/known data
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. Relevant equations

E=kq/r^2

3. The attempt at a solution

8.99e9*46e6/60cm=6.892333333e15

2. Apr 2, 2008

### Snazzy

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 $$\frac{Q'}{Q}=\frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3}$$

or:

$$Q'=Q\frac{r^3}{R^3}$$

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

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

Last edited: Apr 2, 2008