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)
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
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... :(
Homework Statement
∫dt/(t^2 +2tcos a + 1)
(Limits of the integral are from 0 to 1)
(0<a<π)
Homework Equations
The 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...
@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.