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.
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.
Lagrange's mean value theorem
The Attempt at a Solution
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... :(
∫dt/(t^2 +2tcos a + 1)
(Limits of the integral are from 0 to 1)
The Attempt at a Solution
Put t=sin a
∫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...
When I calculated the answer, it came out to be 40.9 cm. Strange!! I used the lensmaker equation given in Wikipedia. These were my results : f= 900/37 cm, u=-60 cm (given) So, using lens formula v = 40.9cm. And yes you need to use the thickness. See the lensmaker's equation in Wikipedia!!!
This is just my doubt and not really a question. We know that speed of light changes upon change in medium of propagation. Its wavelength also changes. But its frequency doesn't change. So, does the colour of light change upon change in the medium of propagation...
hi all! I am not able to understand what's a Lewis acid and a Lewis base. tell me if these are Lewis acids or Lewis bases and tell me the reason. HCl, H2SO4, H3PO4, CO2, HCN.
my approach: first 3 I'm not able to decide. 4th one I thought to be Lewis base as 2lone pairs of electrons left in...
hey guys! we know that velocity of a wave = wavelength x frequency. and speed of light changes with change in optical density. so there must be change in either frequency or wavelength or both. which of these quantities change with change in optical density? and why?
huy there! everyone says that light has wave particle duality. it means light is a wave and also consists of a particle. now, wave is a disturbance in a medium. but, we know that light can travel without a medium. then why do we say it is a wave? (i know that it is a non mechanical wave then why...
i cannot believe what a silly mistake i was doing. i was taking everything in feet and second but was taking acceleration in y direction in metre/second square. now i have solved the problem and got the correct answer. the acceleration should be 33 ft/s^2 (approx.).
first of all what type of image are you talking about. is it real or virtual?
if it is a real image:
(a) do you know where the image lies if the object is at centre of curvature? aren't image and object at the same distance from the pole of mirror. if you exactly know what 'R' is, then you...
ok i have proceeded in this way. just check where am i wrong:-
u (in both x and y directions) = 64/root 2 feet/sec
distance of the ball from goalpost = 40 yards = 120 feet
a (in x direciton)=0
therefore, 120 = 64/root 2 x t + 1/2 x 0 x t^2 => t = 15 x root 2 / 8 second
therefore, s (in y...
well i have a suggestion for you. first, read the question carefully and note down all the bits of information. like in the above case: work done and charge. then think of an equation relating the information and the question. this will help a lot.