Recent content by soccerj17

I tried plugging the numbers I knew into the integral and got E=\int(\frac{k\lambda_{0}x^{2}dx}{0.0361(x+0.19)^{2}} and taking the constants out of the equation I got E=\frac{k\lambda_{0}}{0.0361}\int\frac{x^{2}dx}{(x+0.19)^{2}} with the limits of the integral being x=0 to x=L (or 0.19) Does...

Homework Statement A thin rod extends along the x-axis from x=0 to x=L and carries a line charge density \lambda = \lambda_{0}*(x/L)^{2}, where \lambda_{0} is a constant. If L=0.19 m and \lambda_{0} = 40\muC/m, find the electric field strength at the point x=-0.19 m. (Hint: the resulting...

Alright i got it!

So total charge is the original capacitance (40 pF) times the 3 KV so 120 nC. Then I find voltage by V = Qtotal/Ceq which is C1 + C2 then i use that to find Q by using Q=CV?

[SOLVED] capacitor problem Homework Statement A 40 pF capacitor is charged to 3 kV and then removed from the battery and connected in parallel to an uncharged 70 pF capacitor. What is the new charge on the second capacitor? Answer in units of nC. Homework Equations pF= 1 F x 10^-12...

nevermind, i figured out its just half of the system kinetic energy

[SOLVED] Kinetic Energy of Charged Particle Homework Statement Particles A (of mass m and charge Q) and B (of m and charge 5Q) are released from rest with the distance between them equal to 0.9976 m. If Q=33e-6 C, what is the kinetic energy of particle B at the instant when the particles are...

Thanks! That was the problem. My calculator had given me the answer 0.000006 so I had to convert everything to the -6 power to get the more specific answer.

I tried submitting both positive and negative answers, so i don't think that's the problem.

[SOLVED] electric charge problem Homework Statement Two identical small metal spheres with q1>0 and magnitude of q1>magnitude q2 attract each other with a force of magnitude 74.9mN. They are 2.47 m apart. The spheres are then brought together until they are touching. At this point, the...

Ok I got the answer right now. Thanks!

Actually Force F is 1.2 m from the corner, right?

So then torque gravity would be (28kg)*(9.81m/s2)*(.69/2m)? And Force F is .69 m from the bot right corner right?

So the gravity torque isn't 164.64? I thought torque is force times perpendicular distance to the pivot point. Also I still don't see how to solve the problem. What 2 torques do I equate?

[SOLVED] Torque Homework Help Homework Statement Here's the problem and I need to submit the answer online tonight! Consider the rectangular block of mass m = 28kg, height h = 1.2 m, length l = 0.69 m. A force F is applied horizontally at the upper edge. The acceleration of gravity is 9.8...