Electric Body

Electricity and Magnetism Level 2

A thin, non-conducting hemispherical shell contains a positive charge $Q$ on it, which is uniformly distributed on the shell. Point $\text{O}$ is the center of the base circle, and point $\text{P}$ is on a diameter of the base circle some distance away from $\text{O}.$

Which of the vectors shown below represents the direction of the electric field at point $\text{P}?$

A B C D

9 solutions

Steven Chase
Dec 5, 2017

1) There is obviously a downward component

2) To address the question of a lateral component, consider a fully-closed sphere. The field would be zero in that case. However, both the upper and lower halves of the sphere would contribute identically to the lateral component. Therefore, the lateral contribution from each half must be zero.

Why is the lateral component be zero in case of a sphere?

Sami Emad - 3 years, 5 months ago

Either inside or outside the sphere, the gravitational potential will be a spherically symmetric solution of Laplaces’s equation, and so will be of the form $A+Br^{-1}$ for constants $A,B$ , where $r$ is the distance from the centre of the sphere. Inside the sphere the potential must be continuous at the origin, and so must be constant. Hence the electric field is zero there.

Mark Hennings - 3 years, 5 months ago

At the origin (r=0), r^-1 is undefined. right? How can the potential be continuous when the lim r^-1 as r goes to 0 creates a discontinuity?

Gene Waldenmaier - 3 years, 5 months ago

@Gene Waldenmaier – By choosing $B=0$ , which is what makes the potential constant...

Mark Hennings - 3 years, 5 months ago

Oh yes... i imagined a spherical thing. Forgot that hemisphere term.

Ananya Aaniya - 3 years, 5 months ago

I took a shell and charged it. Then I took a thread and lowered it toward the shell and it moved to the center.

Jim Dorval - 3 years, 5 months ago

Digvijay Singh
Dec 5, 2017

Laszlo Kocsis
Dec 19, 2017

Complete the hemispherical shell with its mirrored copy to get a hollow sphere. Shell theorem says force is zero inside a hollow, spherically symmetric ball. The resultant of the forces exerted by the two hemispherical shells are zero only in case C.

Why? I get that if we complete the hemispherical shell the C vector would be canceled because the resultant force would be zero. But wouldn't this happen with any vector (A, B, C or D)? The way I see it is that if it was a sphere, no matter what direction you draw the vector, the opposite side would still have an identical mirrored vector (because it is a sphere), thus resulting zero anyway. For example, if we draw the vector D at the point P (as the image shows), one could say that there could be an identical opposite vector located at a point in the left side of the diameter (let's call it -P to make visualization easier). By doing this the resultant force at O would still be zero. What I'm saying is that any vector could be drawn at P and the point O would still have a resultant zero provided that we draw the same vector with opposite direction at the opposite side of the sphere to cancel it out. I'm probably missing something very stupid, but I'm not understanding how could someone arrive at a solution if it seems possible to figure out a way to make any of the vectors result zero at the point O.

David Adler - 3 years, 3 months ago

The two half spheres are mirror symmetric, so the directions of their electric fields are mirror symmetric, too. Not point symmetric, as you suggest.

Laszlo Kocsis - 3 years, 3 months ago

Mark Hennings
Dec 18, 2017

For interest, look here for a general solution determining the magnitude as well as the direction of the electric field at $P$ .

Nandana Pandya
Dec 24, 2017

The theory says that...In case of a full sphere the electric field must be zero...Thus only C case has the ability to do that.

Robert Edward
Dec 22, 2017

The right hand thumb rule of the flow of current

Priyanshu Mishra
Jan 7, 2018

Assume to the contrary that the net electric field is not in vertically downward direction, then it must be at some angle with horizontal. Now hold the sphere with your hand and rotate it by 180 degrees in the same plane. Now the electric field at that point will be in the direction opposite to the direction we presumed. But in space there can not be two directions of electric field at any point. Hence our assumption is wrong.

So the electric field must be in vertical direction.

Justin Chang
Dec 20, 2017

Use the newton's shell theorem where it says that the force is 0. Only in C,the force in the two shells is zero.

Sy Ed
Dec 20, 2017

by using right hand rule it can be determined and so it is c only

Could you explain what is the right hand rule? I am not aware of it.

Pranshu Gaba - 3 years, 5 months ago

Right hand rule is for induced magnetic field due to direct current. Put your right thumb parallel to the wire carrying current with the thumb pointing in the direction of current flow and fingers wrapped around the wire. Your fingers show the direction of magnetic field.

Michael Nicholas - 3 years, 5 months ago

Thanks, but I don't see how this rule is relevant is solving this problem. There is no current in this problem.

Pranshu Gaba - 3 years, 5 months ago

its actuaaly right hand thumb rule shows induced emf direction

sy ed - 3 years, 5 months ago

Electric Body

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