- #1

- 54

- 3

- Homework Statement
- See picture.

- Relevant Equations
- Gauss law for polarizaton : ## -Q = \int P dA ## , dA = element of area

Attempt at solution:

a)

**Since I need help with b) this section can be skipped.**Results :

##ρ_{psa} = -Pa ##

##ρ_{psb} = Pb ##

##ρ_{p} = \frac {-1}{R^2} \frac {∂(R^2PR)}{∂R} = -3P ##

b) This is where I am unsure (first time using gauss law for P) so I need some confirmation here:

## \int E ⋅ ds = \frac {Q} {ε_0} = \frac {1} {ε_0} \int -P dA ##

This will become:

## E(R) 4πR^2 = -\frac {1} {ε_0} PR4πR^2 ##

For 0<R<a : E(R) = 0

For a<R<b : ## E(R) = -\frac {1} {ε_0} PR ##

For b<R : E(R) = 0

**The problem here is the solution show**: For a<R<b : ## E(R) = -\frac {1} {ε_0} P ##

Did I go wrong somewhere or is this a typo in the solutions?