ABOUT THAT LAST QUIZ
Calculating the Electric Field
November 13, 2018
Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equilateral triangle, as the drawing shows. The magnitude of each of the charges is 5.0 µC, and the lengths of the sides of the triangle are 4.0 cm. Calculate the magnitude of the Electric Field at point B.
It is suggested that you read the textbook on this topic as you consider the following material. The Electric Field is a critical concept in this course as will the Magnetic Field later in the semester. So the concept of a “field” becomes critical from a conceptual point of view.
There are two ways to consider the forces that act on charge B. The first and seemingly the easiest is to simply recognize that there are two (vector) forces acting on charge B; one from charge A and one from charge C. To calculate the force on B in this way we simply use Coulomb’s law twice.
Coulomb’s Law:
,
where the r vectors are unit vectors pointing from each of the charges pushing or pulling on charge B to the charge B itself (C and A). The units of F are Newtons. This is needed to keep the equation indicating a vector sum. Using the Greek letter sigma () to indicate summation, this is often written as:
,
where the i ranges over all of the charges contributing to the force on charge B.
Notice that in the material above, the charge on B is carried throughout the calculation.
The other approach is via the Electric Field concept where we remove the charge B entirely from the picture and calculate the effect that the other charges have on the space where B was located. After this, we return the charge to the point to calculate the force on it. The beauty of this approach is that we can place ANY charge at point B and calculate the force on it without going through the whole calculation again. The Electric Field at a point (B) caused by the other charges is the same as the force that would be on a one coulomb charge placed at that point but watch the units.
So,
We have replaced QB by a 1 (but we omit the unit of charge at this point and will return it later). The units of E are N/C while the units of F are simply N.
We can now calculate the Electric Field at B in two ways.
FORCEAs we did in class, each force vector is along one of the sides of the triangle and all of the angles that we need to deal with are 60 degrees. The force from A is therefore the same as the force from C since both charges are the same. Call this force simply F Since the two vertical components cancel out, the net force to the left is given by
And F is given by
The Electric field is also to the left and is equal to the force divided by the charge or
Just to check,
F=qE
Or
F=(5x10-5)x(2.8x107)=141N
As expected. / ELECTRIC FIELD
The calculation for E is similar but here there is no charge at point B. As before, the angles of importance are 60 degrees. At point B, the electric field E caused by charge A has magnitude
Again, with the same geometry, there is an E along both lines connecting point B to the sources of the field at A and C. Both components are Ecos(600) and there are two of them (as before) so the net field to the left is
Enet=E=2.81x107 N/C
The force on ANY charge at point B is then simply F=qEnet and this provides the same result obtained in the other column.
The equation for E is therefore:
At this point you may be wondering why we need to mess with E when F seems sufficient. There are many reasons for introducing the Electric field. It exists at a point in space no matter what charges are producing it and in an electromagnetic wave (light, radio) travels through space with no charges in the neighborhood but the E field travels along anyway. It is a real, physical quantity which is far more fundamental than F.
The Electric field can also be calculated using the same equation for points at which there is NO charge. Just add the kQ/d2 terms (as vectors) using all of the charges in the neighborhood.
I hope that this helps.
JBB
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