Recent content by rofldude188

Yeah you're right for the first part I need to plug in t = 0. Sorry I misread I thought z was along the horizontal axis but I now see it's along the vertical axis. So in this case you would find the new radius $$r_z^2 = r^2 + z^2$$ and then find area of that circle with respect to time and go...

a) Calculate the proposed induced emf along the equator of the balloon. (horizontal equator), at the moment indicated above. $$V(t) = V + Ft \implies \frac{4 \pi r^3(t)}{3} = V + Ft \implies r(t) = \sqrt[3]{\frac{3V+3Ft}{4 \pi}}$$ $$\phi = B \pi r^2(t) = B\pi (\frac{3V+3Ft}{4 \pi})^{2/3}$$...

That makes complete sense thank you.

a) Find Electric Field at any point in the dieletric in the terms of the parameters given Making a pillbox Gaussian surface with one end in the conductor where E = 0 and the other end in the dieletric we have that $$\oint D \cdot dS = \rho_s A \implies D = \rho_s \implies E = \frac{Q}{A...

a) Find the electric field in space For r < R (where R is the radius of the gold ball), E = 0 because the gold ball is a conductor. For r > R, let us make a Gaussian surface. $$\int \vec{D} \cdot \vec{dS} = Q_{free} \implies \vec{D} = \frac{Q}{4 \pi r^2}$$ Now this is a bit hand wavy but the...