electrostatics dobtSS....:):)

hmmm thnxx bhaiiya

thnxx bhaiiya but is my method fully wrong

THE DISTANCE x U R TALKING ABOUT IS PERPENDICULAR TO THE AREA OF SQUARE OR IN 2-DIMENSIONAL.

What is wrong in my solution ?
dQ=σ.dS
dQ=(ar)(2Ï€r dr)
∫dQ=2Ï€a ∫₀^r r²dr
Q=2Ï€ar³/3

From Gauss Law
E.A=Q/ε
E(4Ï€r²)=2Ï€ar³/3ε

E=ar/2ε

i have taken σ=aR means a.R ie aRcosθ.......is there i ve mistaken.....

neonee der????????

σ = a r cos x , where x is the polar angle .

Polar angle means the angle made by the field vetor with the z - axis .

Now , the field is wrapped around the sphere in such a way that at the end points of a certain diameter , the field is zero and on either side of the diameter , the field is either negative or positive .

You can think of the sphere as superposition of 2 spheres , one having a +ve charge and another having a - ve charge , each being of same size .

So , the portion at the intersection of the 2 spheres has uniform electric field ,

E = ρ L / 3 ε ₀ , where L is the vector separating the center of the two spheres .

Now , you have to find σ in terms of ρ and r only , otherewise you cannot express the result .

As the field is uniform , hence σ = ρ x ( the width of the intersected portion ) = ρ x d ( suppose )

From the figure , try to prove geometrically that , d = | L | cos x .

Hence , σ = a | R | cos x = ρ | L | cos x

Hence , a | R | = ρ | L |

Substituting the value , we get , E = - a R / 3 ε ₀

The minus sign comes because the field vector is oppositely directed as compared to the z - axis .

P . S ------ Capital letters mean vetors . | K | means magnitude of vetor K . The bigger the " + " and " - " signs are , the bigger is the field there . The pink part of the picture is the distance " d " .

@kaymant sir, i agree with the point that you should try on your own. But what i said was, iff you don't get the answer to your question even after putting a lot of fight, you could refer to the book for the solution. I mean eventually, looking at the solution in the book, and solution of a person who solves the problem here is almost the same thing.

As electric field is in x-direction only, flux through surfaces perpendicular to x-axis exists.

For the surface at x=l, A (vec) = -l² i
For the surface at x=2l, A (vec) = l² i

=> flux thru surface at x=l is A (vec) . E (vec) = -l² i.a√l i
= -al^5/2

=> flux thru surface at x=2l is A (vec) . E (vec) = l² i.a√2l i
= al^5/2√2

So, net flux = al^5/2(√2-1)

9) two large ,identical and parallel conducting plates have surfaces X nd Y ,facing each other .the charge per unit area on X is σ₁ , nd on Y it is σ₂.
a) σ₁ =-σ₂
b) σ₁ =-σ₂ only if charge is given to one plate only.
c) σ₁ =σ₂=0 if equal charges are given to both the plates.
d) σ₁>σ₂ if X is givenmore charge than Y.

one more is right

φ = a(x²+y²)+bz²

\vec{E}=-[2ax\vec{i}+2ay\vec{j}+2bz\vec{k}]

hence \left| \vec{E}\right|=2\sqrt{a^{2}(x^{2}+y^{2})+b^{2}z^{2}}
shape of equipotential surface :
Put \vec{\rho }=x\vec{i}+y\vec{j} or {\rho }^{2}=x^{2}+y^{2}

then equipotential surface has the equation a{\rho }^{2}+bz^{2}=constant=\varphi

a) If a>0 , b>0 then φ>0 and the equation of the equipotential surface is \frac{\rho^{2} }{\varphi /a} +\frac{z^{2}}{\varphi /b} = 1

which is an ellipse in ρ and z co-ordinates .
In three dimensions the surface is an ellipsoid of revolution with semi-axis
\sqrt{\varphi /a},\sqrt{\varphi /a},\sqrt{\varphi /b}.

b) If a>0 , b<0 then φ can be ≥0. If φ > 0 then equation is
\frac{\rho^{2} }{\varphi /a} -\frac{z^{2}}{\varphi /\left| b\right|} = 1

this is a single cavity hyperboloid of revolution abt z-axis. If φ=0 then
a{\rho^{2} } -{z^{2}\left| b\right|} = 0
z=\pm\rho \sqrt{\frac{a}{\left|b \right|}}
is the equation of a right circular cone.
If φ < 0 then equation can be written as
\left|b \right|z^{2}-a\rho ^{2}=\left|\varphi \right|
\frac{z^{2}}{\left|\varphi \right|/\left|b \right|}-\frac{\rho ^{2}}{\left|\phi \right|/a} = 1
this is a two cavty hyperboloid of revolution abt z -axis

thnxx mammay bhaiiya nd ahmand bhaiiya ........ve to think over it
@mammay bhaiiya thtss not in jee na