Find the force of repulsion between the two uniformly charged insulating hemi-spherical shells shown in the above figure.
Radius of larger shell = R
Radius of smaller shell = r
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Log in to reply
∣ G q ∣ is the force on a charge q from a hemisphere only.
Log in to reply
but G[q]=0.5G[2q] right?
Log in to reply
@Wing Yan Yau – Yes, but the force between a hemisphere carrying a charge Q at the point charge q (at the centre of the hemisphere) is ∣ G q ∣ = ∫ 0 2 1 π cos θ 4 π ε 0 R 2 2 π R 2 Q q × 2 π R sin θ × R d θ = 8 π ε 0 R 2 Q q acting along the axis of the hemisphere.
Same solution as provided in the Puzzling problem book
Isn't there any other way to do it @Mark Hennings
Log in to reply
Well obviously, the force could be expressed as a multiple integral over two hemispheres, but I don't fancy that; do you?
actually i am not acquainted with such mathematics right now
Problem Loading...
Note Loading...
Set Loading...
Call the larger hemispherical shell H and the smaller shell K . Imagine a second uniformly charged large hemispherical shell H ′ which, together with H , forms a complete uniformly charged spherical shell with total charge 2 Q . Let F H K be the electrostatic force between H and K , and let F H ′ K be the electrostatic force between H ′ and K . Since the electric field inside a hollow uniformly charged hemisphere is zero, we deduce that F H K + F H ′ K = 0 .
If we now imagine a second uniformly charged small hemisphere K ′ , which, together with K , forms a complete uniformly charged spherical shell of charge 2 q , and let F H K ′ be the electrostatic force between H and K ′ then, by symmetry, F H K ′ = − F H ′ K = F H K . Thus F H K = 2 1 ( F H K + F H K ′ ) , and this last is the electrostatic force between H and the entire smaller spherical sphere. Since the electric field of a uniformly charged spherical shell is, outside that shell, the same as the electric field of a point charge (of the same charge as that on the shell) at the centre of the shell, we deduce that F H K is equal to 2 1 G 2 q = G q , where G u is the electrostatic force between H and a point charge u at the centre of the hemispherical shell H .
It is standard bookwork therefore that ∣ ∣ F H K ∣ ∣ = ∣ G q ∣ = 8 π ε 0 R 2 Q q with the force acting along the line of symmetry of the larger hemispherical shell.