Recent content by arnab321

I kinda can't get how to do that transformation. I can see that W and that with one on its sides inverted, something like v^ will have the same MI as they are still at the same dist. from the axis. but how to know that straightening that v^ into a rod of 4L will still have the same MI?

oh, thanks!

wow nvr knew that this forum also has study material :o well, what i was saying was that, the MI of leftmost rod about the grey line would be M(L sin 45)²/3 = M(L)²/6. this result was already derived from the CM of the rod, so i applied parallel axis theorem to shift that axis to a distance...

Homework Statement A W figure. each of the 4 arms of the W has mass M, length L, and are at right angles. axis is in the plane of W, passing through the middle, vertically. Homework Equations i used the result I=M(L sinθ)²/3 for a rod inclined at angle θ with the axis, and axis...

electric field is (potential) x (dist.) but in this case what's the distance? why does it have to be the distance from centre?

edit: oh I'm sorry I didn't see the battery there before the switch is closed. I got the ans. 1.5.

Well, um... Current through inductor is i=i0(1-e^(-t/tau)). Putting t=0 in the equation, I get i=0.

i think that's the Emf and not the electric field. anyways, option d is wrong (anyways, ur explanation wasnt even giving option d)

Homework Statement A cylinder of radius R has a uniform, time varying mag. field B, (dB/dt < 0). magnitude of electric field at a point P at a distance r (<R) is: a) decreasing with r b) increrasing with r c) not varying with r d) varying as r^-2 Homework Equations The...

Homework Statement Homework Equations afaik, flux in inductor = Li. at t=0, current through L is 0. so change of flux = LΔi = L(i-0)=Li The Attempt at a Solution at t=∞, i through cap. is 0. so, i get the following equations: 5(i1+i2)=20 5(i1+i2)+ 5i2=10 i get...

Well, I get that charges will be now at corners, and the z component will be 1/4th of that of the hemisphere, and it will have x and y components also, since they are no longer canceled off by symmetry, But how to say that x & y components will be equal to the z component? Edit: ok I get...

nope. google it. general expression for q on a hemisphere (shell) is E=\frac{kq}{2r^2} i also derived it by considering rings as elements. im a 12th grader now, but such problems are generally not asked in school level, but only at competitive level. anyways, for figuring out E on 1/8...

integration is always the last option, but in the solution sheet, it was derived from the results of hemispherical shell (vaguely explained)... seems like I'm having problem understanding the direction of E, but in my opinion, kq/2r² should be the resultant of all vectors?

I looked up and arrived at this: E of hemisphere with 4q = 4kq/2r² at center, upwards So, for 1/4 th of it, effective charge is q. shouldn't the E be kq/2r² now, in a direction radially outward?

Thnx, convinced...