Metal sphere of radius R carries total charge Q.What is the force of repulsion between "northern" hemisphere & "southern " hemisphere.

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– Harsh Shrivastava · 2 months, 2 weeks ago

Bro in you sbh binary questions, how to decide the limits of integration?Log in to reply

– Spandan Senapati · 2 months, 2 weeks ago

If your form is something like \(da/dt=f(a)\) then set \(a\) from \(a(o)\) to \(r\).And if you are getting higher powers neglect \(r\)Log in to reply

@Aniswar S K. You might want to see this solution 😃 – Harsh Shrivastava · 2 months, 2 weeks ago

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– Aniswar S K · 2 months, 2 weeks ago

Yeah Definitely! I was seeing the solution of P 140 in 200 more puzzling problems in physics.Log in to reply

– Spandan Senapati · 2 months, 2 weeks ago

Which book..The one by peter gnadig??Does it have the same problem?Log in to reply

– Aniswar S K · 2 months, 2 weeks ago

Yes!Log in to reply

– A E · 2 months, 2 weeks ago

Ya I think I saw the problem in peter gnadig's book, though this doubt was from Griffith's.Log in to reply

– Spandan Senapati · 2 months, 2 weeks ago

Thanks a lot.Log in to reply

– Spandan Senapati · 2 months, 2 weeks ago

To have a calculus free solution.Since field is constant \(F=E\sigma πa^2\).this is clear if we analyse the component that's active so Area can be projected...Like the case we do in calculating force due to say air pressure.Log in to reply

– A E · 2 months, 2 weeks ago

Great !!!! Never thought about it.Log in to reply

@Harsh Shrivastava,@A E .Hey I was solving a problem today and got the idea of solving this problem by some symmetry arguments....Can you plz tell me how to post image here.... – Spandan Senapati · 2 months, 2 weeks ago

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Do the same, and then the link that comes in the problem section's body, copy it over and paste it here!

Hope u got it. – Harsh Shrivastava · 2 months, 2 weeks ago

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– A E · 2 months, 2 weeks ago

No idea, I think harsh can help.Log in to reply

– Spandan Senapati · 2 months, 2 weeks ago

I have added it.....Log in to reply

Hey guyz a doubt problem

In this I found the velocity of A by angular momentum conservation. But for velocity of B, why can't we use momentum conservation in horizontal direction? There's no external force acting on the system and thus velocity of b just before striking should also be -v/2?

Also while conserving energy we will have to write work done by tension on block B thus we can't use energy conservation. – Harsh Shrivastava · 3 months, 4 weeks ago

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– Spandan Senapati · 3 months, 4 weeks ago

Clearly the pulley would be applying some force on the system of the 2 blocks and so momentum conservation isn't true.Also your conserving angular Momentum should have been from the pulley point itself or else from other points this force will have some torque.Log in to reply

Pulley is smooth so it will provide only an normal rxn thus we can't conserve momentum thanks!

So how to proceed, how to conserve energy? – Harsh Shrivastava · 3 months, 4 weeks ago

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– Spandan Senapati · 3 months, 4 weeks ago

See one thing that's quite clear from the problem is that block B will strike first the pulley.A simple analogy is is we observe the accn along horizontal of A and B when the string makes an angle \(\alpha\)with the horizontal.For A its \(Tcos\alpha/m\) and for B its \(T/m\).So B strikes the pulley first than A hits the wall.When B strikes let its velocity be \(v\) so A's velocity has 2 components.Radial velocity is v(as along the same string)and Tangential is \(u\).Then COAM gives \(u=v(o)/2\).Now conserving Energy gives \(v^2+u^2+v^2=v(o)^2\).Put \(u\) and \(v=v(o)√3/8\).B comes to rest now so A's radial velocity diminishes suddenly.Now A undergoes Uniform Circular Motion to hit the wall with its tangential velocity\(=v(o)/2\).Got it?Log in to reply

– Harsh Shrivastava · 3 months, 4 weeks ago

Just one more doubt , work done by tension on B? Why is 0?Log in to reply

– Spandan Senapati · 3 months, 3 weeks ago

Like it won't be zero na.It would be moving and T and displacement are in same direction.We applied the conservation of Energy coz the onlynexternal force on the system isn't dissipative like friction.....Can you find the time when B strikes pulley.Its quite interesting problem.Try it.Log in to reply

\( a = \frac{v_{0} ^2 l^2 }{2(2l-x)^3 }\)

On further solving we can obtain time for collision.

Am I correct? – Harsh Shrivastava · 3 months, 3 weeks ago

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– Spandan Senapati · 3 months, 3 weeks ago

Its correct...I too got the same.Log in to reply

Hey,sorry for disturbing....but their is another simple question I am confused about.

"Q.a system consists of 2 metallic spheres of radii r1 and r2 connected by a thin wire and a switch S .Intially S is open,and spheres carry charges q1 and q2 respectively.If switch S is closed,the potential of sysyem is?"

My answer is coming as k(q1+q2)/2(r1+r2). But the answer is given as k(q1+q2)/(r1+r2). – A E · 3 months, 4 weeks ago

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– Spandan Senapati · 3 months, 4 weeks ago

Yes its correct.After closing the switch both will be at the same potential equal to \(V=(q1+q2/r1+r2)\)Log in to reply

– A E · 3 months, 4 weeks ago

Please try this question( I am really confused about conductors) https://brilliant.org/problems/conducting-shells-are-awesome/?ref_id=1351815Log in to reply

Also sphere A has no charge initially? Problem is not properly worded. – Harsh Shrivastava · 3 months, 4 weeks ago

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– A E · 3 months, 4 weeks ago

I mean initially only shell B has charge.Log in to reply

– A E · 3 months, 4 weeks ago

OK....thanks!!! one last query....how to use method of image charges for a sphere if its not grounded ?Log in to reply

– Spandan Senapati · 3 months, 4 weeks ago

All assumption eventually must satisfy the Boundary conditions like maintain the spheres surface as an equipotential surface,potential at infinity=0.Log in to reply

– A E · 3 months, 4 weeks ago

Can you please elaborate?Log in to reply

– Spandan Senapati · 3 months, 4 weeks ago

Let's say you what to do that for a neutral conductor.Then as usual place the image charge \(-qR/d\) at a distance \(R^2/d\) from centre.But again we need to do two things now maintain the neutrality and equipotential surface of sphere.So this can be easily done by placing an equal amount of positive charge \(+qR/d\) at the centre.Log in to reply

– A E · 3 months, 4 weeks ago

But what if the spherical conductor is charged with a charge Q?Log in to reply

– Spandan Senapati · 3 months, 3 weeks ago

Same reasoning place \(-qR/d\) at a distance \(R^2/d\) from centre and \(Q+qR/d\) at the centre...So the spheres potential is \(V=k(Q+qR/d)/R\).....Log in to reply

– A E · 3 months, 4 weeks ago

Thanks!!!!Log in to reply

As long as charge distribution is uniform, center of charge is center of mass, thus you can 'replace' two hemispheres by by two positive charges of magnitude Q/2 situated at their com and thus find repulsion between them.

Sorry can't post a full solution, bcoz I am on mobile. – Harsh Shrivastava · 3 months, 4 weeks ago

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– Spandan Senapati · 3 months, 4 weeks ago

Could you plz elaborate more.I don't find this to be conclusive.As if that were so then electric field(from COQ method)at a point say symmetrically at a distance x from centre would be different from \(KQ/R^3*x\).And there are some restrictions for this like satisfying the eq \(Q(+)/|R-r"|\)=\(\int(k\rho dV/r\)where \(R\) is the position coordinate of the COQ.For ex-following the image charge theory which you must be acquainted with for a spherical grounded conductor the eq above holds and the centre of charge is at the position of the image charge.Log in to reply

I can't understand your argument about restrictions, can you please elaborate?

And yes my solution needs some rigorous reasonings. – Harsh Shrivastava · 3 months, 4 weeks ago

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– Spandan Senapati · 3 months, 4 weeks ago

The eq just meant that if there is a certain charge distribution in space \(\rho(x,y,z)\) with the total charge say \(Q\) can we claim for the existence of a point having coordinates \(int\rho dV*R\)/\(Q\)(follows from the same reasoning as COM \(\int xdM/M\) such that if we place \(Q\)there then the potential distribution will be same in space as produced by the original system.Log in to reply

– Harsh Shrivastava · 3 months, 4 weeks ago

Oh yes. So is my solution correct? and do you know any alternative?Log in to reply

– Spandan Senapati · 3 months, 4 weeks ago

Although what you said abt external point is correct.I think it doesn't work well for internal points...I am trying to solve by some basics but haven't got anything useful..although a trivial approach would be to work in polar coordinates and use double integrals let's refrain from that.Log in to reply

– A E · 3 months, 4 weeks ago

Thanks!!Log in to reply

@Aniswar S K – A E · 3 months, 4 weeks ago

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@Harsh Shrivastava – A E · 3 months, 4 weeks ago

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@Spandan Senapati – A E · 3 months, 4 weeks ago

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