If the charge density was uniform, wouldn't the potential inside be equal to the potential of the surface? V=1/4pi epsilon * Q/R . But in this case the charge density is not uniform. Also it is a conducting shell so i think the electric field inside should be zero because of gauss law, there is...
My question is what would be the integral boundaries for Vinside and Voutside. When you just integrate sin^2x dx you just get one result. How can you write for inside and for outside?
We have a spherical shell. We want to find the electric potential everywhere. Given charge density σ=k*sinθ, k is constant. Also it is wanted from direct integration. V=1/4piε ∫σ/r da. How can we do this, please help?
I thought because it is sphere da is in the direction of r so da=r^2sinθ dθ...