Well srsly that's quite awesome!Thanks bro you have given some gr8 insights to me in physics problem solving. Thanks again and do keep posting such awesome methods!

@Aniswar S K. You might want to see this solution 😃

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.

@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....

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.

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.

@Harsh Shrivastava
–
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?

@Harsh Shrivastava
–
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.

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).

@Brilliant Member
–
All assumption eventually must satisfy the Boundary conditions like maintain the spheres surface as an equipotential surface,potential at infinity=0.

@Brilliant Member
–
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.

@Brilliant Member
–
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 \(V=k(Q+qR/d)/R\).....

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.

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.

The method I have used holds true as long as the point we are considering lies outside the sphere. Thus your argument of electric field at a distance x from center can't be solved by COQ method bcoz it lies inside one of the hemisphere.

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

And yes my solution needs some rigorous reasonings.

@Harsh Shrivastava
–
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.

@Harsh Shrivastava
–
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.

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## Comments

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

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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 \(r\)

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Well srsly that's quite awesome!Thanks bro you have given some gr8 insights to me in physics problem solving. Thanks again and do keep posting such awesome methods!

@Aniswar S K. You might want to see this solution 😃

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

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Thanks a lot.

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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.

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Great !!!! Never thought about it.

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@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....

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See , you know how to post picture in a problem?

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.

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No idea, I think harsh can help.

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I have added it.....

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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.

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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.

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Yes I have conserved angular momentum from pulley only.

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?

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

On further solving we can obtain time for collision.

Am I correct?

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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).

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

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

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Also sphere A has no charge initially? Problem is not properly worded.

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

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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.

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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.

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The method I have used holds true as long as the point we are considering lies outside the sphere. Thus your argument of electric field at a distance x from center can't be solved by COQ method bcoz it lies inside one of the hemisphere.

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

And yes my solution needs some rigorous reasonings.

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Thanks!!

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@Aniswar S K

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@Harsh Shrivastava

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@Spandan Senapati

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