Question 1
I thought that Question 1 would be pretty simple, yet many people got a wrong answer.
Here, we subtract the colored area in the 1st figure from the colored area in the 2nd figure.
A1=.5*(1*3+11*4)
A2=.5*(8*1+13*3+13*1+18*2)
The answer is A2-A1=24.5 units.
Question2
The basic idea was to take an elemental ring at a distance x from the apex. The common mistake in this question would be in the wrong interpretation of the "width" of the ring.
In the 1st diagram, the small black line is dx, while the small red line is the length we are most interested in. Now, we simply write the equations and use an online integrator to get the answer:
Question 3
The obvious way to start is to find the field inside the charged material. Assume that the charged material is composed of infinitely many 'large' charged plates. Use the fact that the field due to each plate is independent of its distance from the plate. :)
Answer: 750 V
If you find a mistake in the solutions, please leave a comment.
My approach for the 3rd one-
ReplyDeleteAccording to the qs,we can take the system to be a collection of infinite plane 'long' charged sheets between the two plates.
Then,finding electric field at a dist 'x' from the left plate-
E(x)= E1 - E2 ;where E1 is the field due to plates on the left and E2 of the plates on right,considering the vector nature.
Integrating limits:0->x for E1 and x->L for E2
,we get -
E(x)= [2x^2-L^2]K/4e; e-epsilon naught
dv=E.dx
Integrating again,limits 0->L
V(L)=K/4e[2L^3/3-L^3]
Putting L=1/20m and value of K
V=750V viz,independant of area of plates....
So where did i go wrong?
yes you're right.. :) sorry for all the confusion!
ReplyDeleteone small calculation mistake from my side..
but still, this change does not affect the results at all..
for question 1 we can simply divide our polygon into triangles and then simply calculate the area of each triangle by |x1 y1 1|
ReplyDelete|x2 y2 1|
|x3 y3 1| this formula.
take the last part as determinant...
ReplyDeleteI am getting 1500V as the ans.I used gauss's law to calculate the electric field a funtn of x (dist. from the 1st place) and got E as kx^2/2e(on integrating for pd webgetbkx^3/6e).Please tell where the mistake is
ReplyDeleteMaybe you just took the field leaving the gaussian surface.I think gauss law accounts for those fields only which leave the surface,but here an opposite field is entering as well.
ReplyDeletein question -2 , calculating the moment of inertia the red line is the slope of the curve at a given point .... so the red part should be 3x^2 and not root(1+9x^4)......
ReplyDeletetell me if i am wrong !!!!!!!!
@Sumantra: I think that you took a Gaussian surface (say cylindrical) perpendicular to the plate. At one end of the cylinder, the field is E(x) which is what you need to calculate. You also need to know the field at the other end. How did you calculate the field at the other end??
ReplyDelete@Pulkit: entering fields contribute negatively to the flux. But you could extend the aforementioned cylinder so that the "other end" lies outside the plate and remove your confusion.
@Abhishek: The red line is the hypotenuse. The black line is the base which is dx. There should be a vertical line whose length is 3x^2*dx.