Recent content by why12

that sounds right. Is this the most concise way to solve this problem.

\widehat{\theta} = \frac{\widehat{\rho} - \widehat{z}}{\sqrt{2}} and \widehat{r} = \frac{\widehat{\rho} + \widehat{z}}{\sqrt{2}} this gives \widehat{\theta} = \widehat{r} - \frac{\widehat{E_0}}{\sqrt{2}} which gives E = E_0 \frac{\sin(\theta)}{\sqrt{2}}...

where does the 4\pi come from because I thought \epsilon_0=1 . I think my potential is wrong because when I calculate the electric field I get E = E_0 Cos(\theta) (1+2\frac{b^3}{r^3} ) \widehat{r}-E_0 Sin(\theta) (1-\frac{b^3}{r^3} )\widehat{\theta}...

No I took the derivative and thought translating SI to Gaussian made the 1/4pi but that is for k not epsilon. there is no 4 pi term but the solution in the book has a 4pi term which means none of my answers are the same as the book.

\sigma = \frac{3}{4\pi}E_0 Cos(\theta)

this is my potential from the boundary conditions at r=b and r infinity. V = -E_0 Cos(\theta) (r-\frac{b^3}{r^2})

The book doesn't have a solution for the voltage just the E field, and my E field is E = E_0 Cos(\theta) (1+2\frac{b^3}{r^3} ) \widehat{r}-E_0 Sin(\theta) (1-\frac{b^3}{r^3} )\widehat{\theta} with my E-field along z-axis. I get that from doing taking divergence of V and multiplying by -1.

I get a different E field than the one in the solution which I give the problem statement. I am not sure if I am doing it right and the book has a wrong solution. As seen in my attempt I assumed it was the same as a shell of radius R=b and proceeded to solve for the boundary conditions but I did...

Homework Statement An uncharged hollow conducting spherical shell with inner and outer radii a,b respectively is placed into otherwise uniform electric field E. Calculate the induced charge density at a and b and electric field everywhere Solution for E ,r>b E = E_0...