# Electrostatics doubt.

Metal sphere of radius R carries total charge Q.What is the force of repulsion between "northern" hemisphere & "southern " hemisphere. Note by A Former Brilliant Member
4 years, 3 months ago

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Sort by: - 4 years, 1 month ago

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 😃

- 4 years, 1 month 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.

- 4 years, 1 month ago

Thanks a lot.

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Which book..The one by peter gnadig??Does it have the same problem?

- 4 years, 1 month ago

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

- 4 years, 1 month ago

Yes!

- 4 years, 1 month ago

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

- 4 years, 1 month 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 $r$

- 4 years, 1 month ago

Great !!!! Never thought about it.

- 4 years, 1 month ago

- 4 years, 3 months ago

- 4 years, 3 months ago

- 4 years, 3 months ago

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.

- 4 years, 3 months ago

Thanks!!

- 4 years, 3 months 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.

- 4 years, 3 months ago

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.

And yes my solution needs some rigorous reasonings.

- 4 years, 3 months 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.

- 4 years, 3 months ago

Oh yes. So is my solution correct? and do you know any alternative?

- 4 years, 3 months 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.

- 4 years, 3 months ago

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).

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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

- 4 years, 3 months ago

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

- 4 years, 3 months ago

- 4 years, 3 months 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.

- 4 years, 3 months ago

Thanks!!!!

- 4 years, 3 months ago

But what if the spherical conductor is charged with a charge Q?

- 4 years, 3 months 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 $V=k(Q+qR/d)/R$.....

- 4 years, 3 months ago

- 4 years, 3 months ago

What is meaning of common potential? Does it means that V = Va + Vb?

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

- 4 years, 3 months ago

I mean initially only shell B has charge.

- 4 years, 3 months ago

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.

- 4 years, 3 months 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.

- 4 years, 3 months ago

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?

- 4 years, 3 months 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?

- 4 years, 3 months ago

Just one more doubt , work done by tension on B? Why is 0?

- 4 years, 3 months 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.

- 4 years, 3 months ago

I found acc of B

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

On further solving we can obtain time for collision.

Am I correct?

- 4 years, 3 months ago

Its correct...I too got the same.

- 4 years, 3 months ago

@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....

- 4 years, 1 month ago

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.

- 4 years, 1 month ago

No idea, I think harsh can help.

- 4 years, 1 month ago