Why a smaller sphere having a smaller charge can have an electric field that is stronger than a larger sphere having a larger charge?

This is a followup post to my previous post on the misconception that a smaller spherical conductor will have a larger charge and hence there is a greater possibility of it being discharging.

I did the same question with another of my class today.  However, this time, this class of students had more problems in visualising why a smaller sphere having a smaller charge could possibily have a stronger electric field at the surface.  I also had students who told me that their secondary school teachers told them that it was the charge per unit area that mattered and asked me how to reconcile this with the concept of electric field being stronger.

Let us address the first issue first.  To aid in the visualisation.  I drew the following picture.  In this case, we had already made a calculation of 0.3 micro-coulomb on the small sphere and 0.9 micro-coulomb residing on the larger sphere.  So if there is one electric field line drawn for each 0.1 micro-coulomb charge on it sphere, then we see can have:

So from the diagram above, we see that the although there is less charge on the smaller sphere, the electric field lines could be closer, and hence the E-field could be stronger.

Now for the second question, was the secondary teacher right in saying that it is the charge per unit area that mattered.  Two ways to understand this, first the diagram shows fundamentally an essential point in drawing field lines, the no. of field lines are proportional to the charge, therefore have a greater no. of charge per unit area, essentially means more electric field lines per unit area and hence the field lines will be closer and in other word, the electric field is stronger.

Alternatively, those who prefer to see equations will have 

From the equation, we can see that the electric field strength at the surface of the charged sphere can be written as a constant multiplied by  (Q / surface area of sphere), so essentially the secondary teacher who relates the electric field at the surface to the Q per unit area is actually correct.

A Common Misconception: More Charges Accumulates on a Sharper Point...Is it Really True?

[Picture Source: stock.xchng, "Raw Power1" (2008) by gun4fire]

When I was an O-level student, I was often told by my Physics teacher that as more charges tend to accumulate on a point that is sharper and hence there is a greater likelyhood for discharge and therefore, lightning rod are built to be sharper.  As a typical O-level student, I was like my student now, and just readily accepted what my teacher said then.

While teaching electric field at A-levels, we have a typical problem whereby we have two spherical conductors in which we connect a "long" wire across them.  We then place a certain amount of charge onto the system and the charges with redistribute till they come to electrostatic equilibrium.  (A similar problem can be found in Serway's "Physics for Scientists and Engineers"(6th Edition), Problem 25.50, pg 791.)

After the calculation, we may find some interesting results.  The electrical charges collected on the larger sphere are in reality more than that of the smaller sphere.  However the electric field at the surface of the smaller sphere is large.  The electric field represents the force acting on per unit charge, and hence logically the larger the force the higher the probability of discharge.  Hence I guess, the idea of more charges accumulating at a sharper point and hence greater probability of discharging, can be quite misleading....


PS:  This article is a followup on a previous comment contributed which triggered this idea.  It is possible to start off with the Windhurst machine activity and then follow up with the problem above and a discussion of breakdown voltage.

Problem 25.50 in Serway (6th Edition), pg 791:
Electric charge can accumulate on an airplane in flight.  You may have observed needle-shaped metal extensions on the wing tips and tail of an airplane.  Their purpose is to allow charge to leak off before much of it accumulates.  The electric field around the needle is much larger than the field around the body of the airplane, and can become large enough to produce dielectric breakdown of the air, discharging the airplane.  To model this process, assume that two charged spherical conductors are connected by a long conducting wire, and a charge of 1.20 micro-coulombs is placed on the combination.  One sphere, representing the body of the airplane, has a radius of 6.00 cm, and the other, representing the tip of the needle, has a radius of 2.00 cm.  (a)  What is the electric potential of each sphere?  (b)  What is the electric field at the surface of each sphere?

Difference Between Electric Field and Electric Field Strength

To put it simply, the electric field is a region of space in which when a charged object is placed in the field.  It will experience a force acting on it.

However, in Physics, we would prefer a more specific way to define a quantity, and as with gravitational field, we would like to have a definition that allows us to quantify a physics physical quantity.  Hence, we assign an number (and also a direction) to every point in the field.  In this way, we can find out about the force acting on the charge placed at the point in the field if we were to know the charge and the field strength.

The electric field strength is a property of the field and not charge placed there.  Hence, we define the electric field strength at a point in an electric field as the force acting per unit charge on a small positive test charge when placed at that point in the field

Electric field strength at a point in field, E =  F/q

where F is the force acting on the charge q placed at that point in the electric field.

This equation not only defines for us the magnitude of the field strength at a point, but also the direction of the field.  Although the definition specifies a positive test charge, in practical test of the field, we can always use a negative test charge, the q will then be negative.  What it simply means is the field would be in opposite in direction to the force experienced by the negative test charge.

Students also often ask why there is a need to have a small test charge.  This is to ensure the that the test charge itself does not distort the original field significantly.