Recent content by gnrlies00

potential on sphere of radius R V=-kQ∫∞R1/r2dr = kQ/R ----------------(1) potential on spherical shell is V= 4kQ/5R -----------------(2) (2)-(1) V=kQ/5R--------------(3) E=-∇V=kQ/5R2 ⇒W=εo/2 ∫R5RE2dτ ⇒W=εo/2 (kQ/5)2∫R5R 1/R2 4π dr = 2kQ2/5R ----------------(4) Potential on sphere of radius...

okay, does that mean I should find the potential difference between the sphere of radius R and spherical shell, and find the Energy. then the potential of expanded sphere of radius 3R and spherical shell, find the energy. and Difference in energy is the work done?

Homework Statement Consider a non-conducting sphere of radius 'R' and charge 'Q' is enclosed by spherical shell of radius '5R' and charge '4Q'. If inner sphere is expanded to radius '3R'.Then amount of work done by the field in this process is Homework Equations W=ε0/2∫E2dτ The Attempt at a...

Thanks guys! appreciate your help...

I did that in my previous post and I got the answer 24 meters, but the correct answer is required to be 16 meters.

P1 is the y-coordinate where the peacock catches the snake, in this case P1=0 therefore 36-S1=12-0 S1=24 But on the book they have given the answer as 16 meters.

the distance are equal \Rightarrow 36-S1=12-P1 \Rightarrow S1-P1=24 Is this correct? If yes, then how do I get the second equation? Please help...

yes, I have drawn the sketch. and I know that this problem is very easy, but I can't figure out how to go about the solution. Please give me some hints.

Homework Statement A peacock perched on top of a 12m high tree spots a snake moving towards its hole at the base of the tree from a distance equal to thrice the height of the tree. The peacock flies towards the snake in a straight line and they move at the same speed. At what distance from the...

oops, f(∏)≈e-2∏ (0+i) ≈ ie-2∏

does the limit exists for lim {\sum\frac{1}{m} - ln N}; as N→∞, Ʃ is from m=1 to N Please give me hints, on how to go about this problem thanks

the proof to Euler's formula eiθ=cosθ+isinθ is using taylor series for the exponential function For any complex number z, we define ez by ez=\sum\frac{z^n}{n!}.

Thank you guys for all your help, writing √(3-2i) interms of u+vi, we get (1.81-0.55i), which I had previously rounded up into (2-0.5i) Hence after applying the conditions, the general solution is f(x)=e-(1.81-0.55i)x which is approximately f(x)≈e-(2-0.5i)x \Rightarrow...

could you please explain how to go about, so as to get (2 - 0.5i)2 also, wrt the above quote, my general equation now become f(x)= e3.75-2i

the general solution is f(x)= e√(3-2i)x Converting z = √(3 - 2i) into polar form I get z =± r1/2 eiθ/2 = ±[131/4 e163i/2] ⇒ z=-2+0.5i or 2-0.5i the general solution becomes ⇒ f(x)= e(-2+0.5i)x + e(2-0.5i)x Is this correct??