Recent content by careman

If P(x,0,z) is a point on the xz plane, then the electric field on P due to the two charges, would be the vectorial sum of the electric fields due to each charge, i.e. \vec{E} = \frac{q \hat{r_1}}{4 \pi \epsilon_0 r_1^2}+\frac{q \hat{r_2}}{4 \pi \epsilon_0 r_2^2} = \frac{- q \hat{j}}{2...

Well, yes the electric field is time changing, so I assume I can't use Gauss's law. How can I express the electric field as a function of time?

The displacement current will also be zero? (because from Gauss's law the surface integral of E.dS will be zero?). Therefore the magnetic field is zero?

Homework Statement Two equal and opposite charges (± q) start from positions ± y, respectively and move on the y-axis with constant velocity u, so that they approach symmetrically the axis origin. Using generalized Ampere's law find the magnetic field in any given point at the xz plane...

Thank you for your help and for checking my work. I really appreciate it.

I found the angular velocity ω from the conservation of energy: \frac{1}{2}Iω^2=Mgh ω=\sqrt{\frac{3g(1-cosθ)}{L}}=2.35 rad/s The angular momentum of the sign is L_s=\frac{1}{3}ML^2=0.47 Nms while the angular momentum of the snowball is L_b=mvL=0.32 Nms Comparing these two, the angular...

Thank you for your reply. The moment of inertia is I=\frac{1}{3}ML^2, therefore the angular momentum of the sign when it passes the vertical is L_s=Iω=(\frac{1}{3}ML^2)ω.

Homework Statement A thin uniform rectangular sign hangs vertically above the door of a shop. The sign is hinged to a stationary horizontal rod along its top edge. The mass of the sign is 2.40 kg and its vertical dimension is 50.0 cm. The sign is swinging without friction, becoming a...