OA 
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@albiesriram said:
Plse solve urgent
Determining how the charges distribute on the surface of a conductor is, in general, a very difficult problem. We know that if we charge a conductor the charges go to the surface and redistribute so that the electric field in the conductor vanishes. One of the few shapes for which this distribution can be determined analytically is the ellipsoid.
x^2/a^2 + y^2/b^2 + z^2/c^2=1.
Here, a, b and c are the ellipsoid's semi-axes. One can prove that for an ellipsoidal conductor the surface charge density is given by
σ(x,y,z)=Q/(4 pi abc) X 1/root(x^2/a^4 + y^2/b^4 + z^2/c^4)
where Q is the the net charge of the conductor. Note that if we set a=b=c=R we obtain the uniform charge distribution Q/(4 pi R^2), corresponding to a spherical conductor. Suppose that we measure the electric field near the surface of a charged ellipsoid with Q=1nC, a=2cm,b=5cmand c=3cm. What is the maximum value in in volts per meter of the electric field?
Details and assumptions
k=1/(4 pi ϵ0) =9×10^9m/F
x^2/a^2 + y^2/b^2 + z^2/c^2=1.
Here, a, b and c are the ellipsoid's semi-axes. One can prove that for an ellipsoidal conductor the surface charge density is given by
σ(x,y,z)=Q/(4 pi abc) X 1/root(x^2/a^4 + y^2/b^4 + z^2/c^4)
where Q is the the net charge of the conductor. Note that if we set a=b=c=R we obtain the uniform charge distribution Q/(4 pi R^2), corresponding to a spherical conductor. Suppose that we measure the electric field near the surface of a charged ellipsoid with Q=1nC, a=2cm,b=5cmand c=3cm. What is the maximum value in in volts per meter of the electric field?
Details and assumptions
k=1/(4 pi ϵ0) =9×10^9m/F
@albiesriram said:
Extend 2 sides and form an equilateral triangle of side 7.
Then it is a cakewake to find the ratio as 7/12
Then it is a cakewake to find the ratio as 7/12
@albiesriram said:
Assume K = -1
Now let AB =1
So BC = 1
AC = root(2)
CD = root(2)
AD = 2 = DE
AE = root(8)
So (AE/BC)^2 = 8
Using options (c) seems good
Now let AB =1
So BC = 1
AC = root(2)
CD = root(2)
AD = 2 = DE
AE = root(8)
So (AE/BC)^2 = 8
Using options (c) seems good
@Tusharrr said:Plse solve urgentDetermining how the charges distribute on the surface of a conductor is, in general, a very difficult problem. We know that if we charge a conductor the charges go to the
Dude, this problem is not deserved to be there in CAT. just imagine there will be non engineers. They wont find it doable.. Even non core students other than EC and EE will find it difficult.. And above all it uses differentials nd integrals, laplacian to be precise which is far from the scope of CAT.
Anyway electric field at which point on the surface of the conductor?? Even components were not mentioned..l
Even components were not mentioned..lye kya h yaaar :)
3
@Tusharrr said:Plse solve urgentDetermining how the charges distribute on the surface of a conductor is, in general, a very difficult problem. We know that if we charge a conductor the charges go to the surface and redistribute so that the electric field in the conductor vanishes. One of the few shapes for which this distribution can be determined analytically is the ellipsoid.x^2/a^2 + y^2/b^2 + z^2/c^2=1.Here, a, b and c are the ellipsoid's semi-axes. One can prove that for an ellipsoidal conductor the surface charge density is given byσ(x,y,z)=Q/(4 pi abc) X 1/root(x^2/a^4 + y^2/b^4 + z^2/c^4)where Q is the the net charge of the conductor. Note that if we set a=b=c=R we obtain the uniform charge distribution Q/(4 pi R^2), corresponding to a spherical conductor. Suppose that we measure the electric field near the surface of a charged ellipsoid with Q=1nC, a=2cm,b=5cmand c=3cm. What is the maximum value in in volts per meter of the electric field?Details and assumptionsk=1/(4 pi ϵ0) =9×10^9m/F
not sure abt the units but it is arnd 8.205 X 10 pow (something)
it is a max/min qsn...
x2/a2 = y2/b2 = z2/c2 = 1/3
sigma = 1 X (9 X 10^9/30) X 1/root(1/3 X (1/4 + 1/9 + 1/25))
uske baad calci use karke- 8.205 , order of magnitude aur units ke baare mein sure nahi hu..
@2013IsMine said:x^3+bx^2+cx+db+c+d=4 from 14b+2c+d=0 from 29b+3c+d=-16 from 3subtarcting 1 and 23b + c =-4 ....4subtracting 2 and 35b +c =-16.........5subtracting 4 and 52b=-12...b=-6c=14d=-4so f(x)= x^3-6x^2+14x-4f(4)= 64-96+56-4=64-40-4=20??
why u took equation as x^3+bx^2+cx+d it could be ax^3+bx^2+cx+d
@vbhvgupta said:why u took equation as x^3+bx^2+cx+d it could be ax^3+bx^2+cx+d
coefficent of x^3 was given 1 right in the question?
IF the eqn x^5 + 15x^4 + 85x^3 + 225x^2 + 274x +a - 119 = 0 has exactly 5 negative roots then the value of a can be
100
85
120
90
@vbhvgupta said:IF the eqnx^5 + 15x^4 + 85x^3 + 225x^2 + 274x +a - 119 = 0 has exactly 5 negative roots then the value of a can be1008512090
it can be 120 :)