Recent content by allison west

Okay, so ##V(x) = -15x + V_0## when ##x \lt -1## making ##\vec E_x = 15 \hat x##. ##\rho## in the outer region would be zero because it's outside of the slab. I got ##\rho## inside the region by using Gauss' Law: ##\rho = \frac Q V## ##\frac {q_{encl}} {\epsilon _0} = \oint \vec E \cdot d...

So would it be correct to say that ##V(x) = -15 x## outside of the bounds? When I input the values of ##E_x## I would get: outside of the bounds: ##\rho = \frac {-15 \epsilon _0} {x}## inside of the bounds: ##\rho = -15 {\epsilon _0} ## Or do I have to differentiate ##E_x## in Gauss' law?

Yes sorry, ##V(x) = \frac {15}{2} x - \frac {25}{2}## Yes, I think it would be scalar potential. I got ##V(x)## from the plot which I guessed given the coordinates of a couple points and the fact that I knew it was linear. It looks like one point on the graph is (10, 2) and that the...

Homework Statement A semi-infinite (infinite in y and z, bounded in x) slab of charges carries a charge per unit volume ##\rho##. Electric potential due to this slab is a function of horizontal distance, x from the center of the slab. It is linear for ## x \lt -1m## & ## x \gt 1m##, and is...

Thanks for the help!

Okay that seems to be what I was missing, I was not inputting ##z_0## back into my definition for ##\vec r##. By putting that into the equation I got: ##\vec A_{dip} (\vec r) = (\frac { \mu_0 m_0 } {4 \pi z_0^2})(sin(\omega t) \hat x - cos(\omega t) \hat y)## and ##\vec B = \vec \nabla...

Homework Statement A rotating magnetic dipole is built by two oscillating magnetic dipole moments, one along the y-axis and one along the x-axis. Find the vector potential at a point: (0, 0, ##z_0##) along the z-axis. Then find the magnetic field at ##z_0## . As the magnetic field is a function...

Hi I'm an undergraduate student from the University of Lethbridge in my 3rd year of a physics degree.