Recent content by utkarsh009

This Q is actually the charge on the outer surface (d) . For inner surface (c), it will be negative of the charge on solid sphere. Add both of these to get the net charge. That will be the answer to (b)

v=√(T/¢) T= tension in wire ¢= mass per unit length

Go ahead and use it. See if you can get the answer. (∫E.dS=q/ε(o)) or you could simply use the formula you mentioned it in relevant equations. I'm not supposed to spoonfeed you. Check the answer in your textbook.

Assume that the solid sphere and induced charges are not present at all (since the have no effect). Only a spherical conductor is present with charge on its outer surface.

Your answer to b part is wrong. There will be induced charge on the inner surface of the shell so that the field due to solid sphere and induced charge is 0 everywhere outside it. So the field at 50 cm is only due to the charge on outer surface. Use it to calculate net charge on shell.

What if you integrate it from n to n+1? Also can you think of the area of rectangle formed with height 1/(n+1) and width (n+1)-n? Just compare them on the graph of f(x)=1/x

Yah... Was just thinking the same and was about to post it... :P

But your h(x) had solved the problem. What was the error??

Thanks mate! Solved it... But how did you think of that function h(x)??

Homework Statement Homework Equations Lagrange's mean value theorem The Attempt at a Solution Applying LMVT, There exists c belonging to (0,1) which satisfies f'(c) = f(1)-f(0)/1 = -f(0) But this gets me nowhere close to the options... :(

Oh yah... Thanks ... I solved it... But have a look at this file... I couldn't get an answer for this...

Oh yes... By mistake i typed the integration symbol after it. Cosec a should be inside the integral. So, how should i proceed??

Homework Statement ∫dt/(t^2 +2tcos a + 1) (Limits of the integral are from 0 to 1) (0<a<π) Homework EquationsThe Attempt at a Solution Put t=sin a dt=cosa da ∫dt/(t^2 +2tcos a + 1) = ∫cos a da/(sin^2 a + sin 2a + 1) [ limits of integration changed to 0 to π/2] = ((cosec a)/2) ∫sin 2a da/(sin^2...

Thanks! www.physicsforums.com/showthread.php?t=459065 this thread also helped me. Thanks for giving the link.

@Danny :the equation which you are using seems to be derived from thin lens equation where you neglect thickness as it's too small compared to Radii of curvature.