Recent content by 123yt

Homework Statement Consider a horizontal square plate capacitor of area 1*1 m2, capacitance in vacuum 2 uF, which contains a dielectric material with dielectric constant K=5. The dielectric slides frictionlessly and is attached via a massless string and a massless pulley to a block of mass 2.5...

5 pictures, 5 problems. Fourth one is the fourth picture. My book says I'm supposed to point my thumb in the direction of the current and look at how the fingers curl for magnetic field instead. Using what you just told me, I can get the solution to the fifth problem, but it doesn't help me...

I did. Every single source I've read tells me I need two of the three vectors to find the third. The first one only gives me current, the second one doesn't give me anything, I just realized I can actually figure out the third one, the fourth one only gives me current, and the fifth one only...

How do I find the direction of these things? The right hand rule only works when I have the current and it's asking for clockwise/counter-clockwise, and the left hand rule only works when I have two of the three: Current, Magnetic Field, and Force. Example...

Alright, it works now. Thanks for the help.

Ok, and the integration of 4xydy from 0 to 0.02, and then multiplying that result by x is 0.000016. emf = d/dt * t^2 * 0.000016. Deriving gets 0.000032 * t Plug 1.5 into t to get 0.000048 V. Was the answer supposed to be positive?

1. In the figure the square loop of wire has sides of length 2 cm. A magnetic field points out of the page; its magnitude is given by B = 4.0 t^2y where B is in teslas, t is in seconds and y is in meters. Determine the emf around the square at t = 1.50 s...

Yeah, sorry, copy and pasted the question weirdly. Anyways, thanks for the correction. I got it figured out now.

Homework Statement A long straight wire with circular cross-section and a diameter of 5.10 mm is made of a metal with resistivity = 5.00×10-8 m. The wire carries a current of 28.5 A. Calculate the current density in the wire. Homework Equations J = i / A The Attempt at a Solution...

So I just add the two electric potentials together instead of subtracting them? Could you explain why they add together instead of canceling? If I released the particle at the center, there wouldn't be any net force exerted on it, and it would remain stationary.

If the charge was on the opposite side of the point, the electric field lines would be pointing in the opposite direction, so wouldn't the opposite E-field lines become negative for potential and cancel out the other charge?

Shouldn't it have the exact same value as the potential at infinity?

Homework Statement Two equal point charges Q = 5.18 C are separated by a distance d = 2.00 m. (See figure.) http://homework.phyast.pitt.edu/res/msu/nagytibo/Electromagnetism/Electrostatics/graphics/010a.gif Point A is halfway between the charges and point B is located 1.00 m to the...

Alright... That's somewhat counter-intuitive for me. So does that mean that if I move a charge from infinity to the middle of the two opposite charges, no net work would be done?

Why is the electric potential in the middle of two oppositely charged points 0? Does this also mean that there's no potential energy at this point?