Archives of Ask A Scientist!

About "Ask A Scientist!"

On September 17th, 1998 the Ithaca Journal ran its first "Ask A Scientist!" article in which Professor Neil Ashcroft , who was then the director of CCMR, answered the question "What is Jupiter made of?" Since then, we have received over 1,000 questions from students and adults from all over the world. Select questions are answered weekly and published in the Ithaca Journal and on our web site. "Ask A Scientist!" reaches more than 21,000 Central New York residents through the Ithaca Journal and countless others around the world throught the "Ask a Scientist!" web site.

Across disciplines and across the state, from Nobel Prize winning scientist David Lee to notable science education advocate Bill Nye, researchers and scientists have been called on to respond to these questions. For more than seven years, kids - and a few adults - have been submitting their queries to find out the answer to life's everyday questions.

Previous Week's Question Published: 22 September, 2008 Next Week's Question
Why a force behaves a certain way in an empty charged sphere
Question
In an empty charged sphere, even though the charge is evenly distributed on the surface, why does it behave as if it is all in the center? (The same as gravity)

Question
What you want to find is the strength of the force that the charged sphere would exert on another (unit) "testcharge" somewhere outside the sphere( this is also called the electric field generated by the charged sphere). In order to find out how big that force is imagine for a moment that you are the testcharge walking around the sphere (such that you are always the same distance from the center). Whenever you look back at the sphere, it will always look the same to you. And the force acting on you from the sphere would also have to be the same (see Fig.1).

Figure 1: The electric fields at points A and B should have the same strength, because from the point of view of someone at point A or point B the charges on the sphere look the same. Also, the force should point radially outward. If it didn't, which one of the two dashed arrows would it rather choose at point A?

You can also see that the force acting on you has to be pointing radially outward: imagine again that you are looking back at the sphere. There is nothing that would distinguish the direction to the left or right. So if the force were not pointing radially outward, where would it point? Would it be tilted to the left or right of your position? Since the sphere looks the same when you look to the left or right, the force from the sphere should not distinguish between these directions either (see again Fig. 1). Physicists would use the jargon "since the charge distribution is spherically symmetric, the electric field has to be radial and also spherically symmetric," which means the following: if you rotate the sphere around its center, the charges will remain at the same place, so the force generated by the charges on the sphere should not change either, since it corresponds to the same source. This can only happen if the electric field is pointing radially outward and the strength is uniform at a given distance from the center. Note, that until now I have in no way referred to any of the laws of electricity, only used the very convenient distribution of the charges. In order to actually determine the strength of the force, I do need to use one of the basic laws of electricity, usually referred to as Gauss's Law. For the kind of spherically symmetric setup we have here, you can phrase Gauss's Law in the following way: surround the center of the sphere with another imaginary sphere of arbitrary size. The strength of the electric field at the surface of this second imaginary sphere will only depend on the total amount of charge contained inside this imaginary sphere. Using this basic law it is easy to find what the field will be: as long as the imaginary sphere is within the real sphere there is no charge contained at all, and the electric field will be zero. Once the imaginary sphere is bigger than our real sphere the contained total charge is always the same, and it does not matter where exactly you put it: all of it at the center of the sphere as a point charge, or any other way distributed (as long as it is still spherically symmetric). This is illustrated in Fig.2.

Figure 2: The strength of the field is determined by how much charge is contained within the imaginary (dashed) sphere. As long as we are inside the sphere no charge is contained, so there is no electric field. As soon as you get outside the total surface, charge is contained in the imaginary sphere, and you get the electric field to be equal to that of a point charge at the center.

It is then easiest to pick the case when all of the charge is at the center, and so you see that the field has to agree with that of a point charge at the center of the sphere. I should emphasize once more that this is only true as long as you are outside the sphere; inside the electric field will be vanishing. The final field configuration is illustrated in Fig. 3.

If electric charges distributed on a sphere are too obscure for you, you could use very similar arguments regarding the gravitational force of the Earth exerted on a rocket: the force will be as if the whole mass of the Earth was concentrated at the center of the Earth.

Figure 3: The final field configuration.