What is the field outside the shell?

[tandoorichicken]

A point charge Q is at the center of a conducting spherical shell of radius R. The total charge of the shell is -Q. (a) What is the field in the region between the point charge and the shell? (b)What is the field outside the shell?

I think I got part (a): $$F = \frac{kq_1 q_2}{r^2} = \frac{kQ(-Q)}{R^2} = -\frac{kQ^2}{R^2}$$
$$E = \frac{F}{Q} = -\frac{kQ}{R^2}$$

Not quite sure how to do part (b) though.

 

[cookiemonster]

Try it using Gauss's Law.

$$\int_S \boldsymbol{E}\cdot d\boldsymbol{A} = \frac{Q}{\epsilon_0}$$

What's the total charge enclosed by a surface surrounding both the shell and the point charge?

On a similar account, I suggest having another look at part (a). Try it with Gauss's Law! Gauss makes life easy, not hard.

cookiemonster

 

[M98Ranger]

The field outside the shell can be found using Gauss's Law, which states that the electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space. In this case, the enclosed charge is still -Q, and the permittivity of free space is a constant denoted by ε0. Therefore, the electric flux outside the shell can be expressed as:

Φ = Q/ε0

Since the electric flux is also equal to the surface integral of the electric field over the closed surface, we can set up the following equation:

Φ = ∫E•dA = E∫dA = E(4πR^2)

Solving for E, we get:

E = Q/(4πε0R^2)

This means that the electric field outside the shell is directly proportional to the charge Q and inversely proportional to the square of the distance from the center of the shell. It also does not depend on the radius of the shell itself, as long as the point charge is at the center.

In summary, the field outside the shell is given by the equation E = Q/(4πε0R^2), where Q is the charge of the point charge and R is the distance from the center of the shell. This field is always directed away from the shell, as the charge on the shell is negative and will repel any positive charges outside the shell.