Recent content by r34racer01

Ok I finally made some progress here. For part a the answer is Φ = E*L^3. I kinda understand why its L^3 but I could use some clearing up. The answer is the same for part b. Part c is where I'm stuck now. I used the formula Q = Φ/epsilon = 7.447e11 but that's wrong. Can someone explain what went...

I'm not quite sure what you mean. At z = 0, 3 of the cube's sides intersect, so I'm not sure which side you're referring to.

For z = 0, that's the origin so I couldn't say. z = 1.3, I'd say E is perpendicular to the area x = 0, again can't say. x = 1.3, I'd say that's also perpendicular to E. y = 0, again can't say. y = 1.3, that would be parallel, no? I can see now why u might be confused by my logic, I think...

What i meant to say is that for pt a at least the field will be parallel to 2 side of the cube, the sides that are parallel to the yz-plane I believe. The field will be perpendicular to the other 4 sides so the flux there should be zero. The flux through the other 2 sides parallel will be equal...

A cube, placed in a region of electric field, is oriented such that one of its corners is at the origin and its edges are parallel to the x, y, and z axes as shown to the left. The length of each of its edges is 1.3 m. (a) What is the flux through the cube if the electric field is given by E...

A thin rod of length L has a non-uniform charge per unit length λ(x) given by λ(x) = A x2 where A = 3 µC/m3 and x is measured in meters from the origin. (a) Find the net charge on the rod for L = 2.9 m. Q = 2.439e-5 (b) Find the net x-, y-, and z-components of the electric field at...

Ok well like I said before I tried F = ma => 0.00918 = 0.003kg * a => a = 3.06 m/s^2. I then did x = x0+v0*t + 0.5*a*t^2 ==> x = 0 + 25(6.5) + 0.5(3.06)(6.5)^2 = 227.1425. I then resolved this into x and y components and got x = 205.86 and y = 95.9. These didn't work so I'm still lost.

Throughout space there is a uniform electric field in the -y direction of strength E = 540 N/C. There is no gravity. At t = 0, a particle with mass m = 3 g and charge q = -17 µC is at the origin moving with a velocity v0 = 25 m/s at an angle θ = 25° above the x-axis. (a) What is the magnitude...

Oh man I can't believe I made that mistake, that means I had the answer days ago. Thanks you so much Doc Al, I would have never noticed that, the answer turned out to be 1.4543N if anyone was wondering.

Two charges Qc and -Qc (Qc = 7 µC) are fixed on the x-axis at x = -6 cm and x = 6 cm, respectively. A third charge Qb = 1 µC is fixed at the origin. A particle with charge q = 0.3 µC and mass m = 6 g is placed on the y-axis at y = 10 cm and released. There is no gravity. (a) Calculate the...

Ok so then it should be -(mg(2δL) + 1/2*k(2δL)^2 + 1/2mv^2) = 1/2m(vmax)^2 => -2(9.81)(0.98) + .5(40)(.98^2) + .5(2)(2^2) = .5(2)(vmax^2) => vmax = 1.99 but that's wrong too.

Ok so are you saying that: (mg(2δL) + 1/2*k(2δL)^2 + 1/2mv^2) = 1/2m(vmax)^2, cause when I solve that I get 6.51 which is wrong.

A spring with spring constant k = 40 N/m and unstretched length of L0 is attached to the ceiling. A block of mass m = 2 kg is hung gently on the end of the spring. a) How far does the spring stretch? dL = 0.49 Now the block is pulled down until the total amount the spring is stretched is...

Yes it is.

Sorry my bad, I've added the rest of the info after part a.), I've also added a link to the picture included.