Recitation Activity 3: Electric Fields from Charge Distributions

Physics 212 - Spring 2000

Recitation Activity 3: Electric Fields from Charge Distributions

NAME: Solution REC. SECTION: ______

ACTIVITY PARTNERS: ______

______

INSTRUCTOR: Redwing DATE: January 19, 2000

This activity is based on the following concepts:

· Electric field is defined as the electrostatic force on +1 C of charge; note that it is a vector and that it is measured in units of N/C.

· To calculate the electric field from many charges, we use SUPERPOSITION:

· If we have a discrete collection of point charges, figure out the electric field vector from each charge using Coulomb's Law and then add all the vectors.

· If we have a continuous distribution of charge, we divide up the distribution into "differential" elements of charge, figure out the electric field from a typical element and then use an integral to sum up all such vectors.

Exercise 1: Electric field from point charges.

The figure below shows 4 point charges located on a circle of radius R. A small test charge - q and mass m is released from rest at the center of the circle. You are asked to determine the acceleration of this particle as soon as it is released. Do the problem in the following steps.

(a) In the figure, sketch four vectors that represent the contributions of each of the 4 charges to the electric field at the center of the circle. See figure

(b) What is the sum of the x-components of these 4 vectors?

kQ/R2 + k(2Q)/R2 = 3kQ/R2

-kQ/R2 - k(2Q)/R2 = -3kQ/R2

(d) Next, use the definition of electric field to determine the magnitude of the total force F on the test charge - q and the acceleration a of the test charge when it is released.

e) Finally, in the figure sketch the path followed by the test charge after it is released.

See dotted arrow in figure

Exercise 2: Electric Field from a continuous charge distribution

Two curved plastic rods, one of charge +q and the other charge -2q, form a circle of radius R in an x-y plane as shown below. The charge is distributed uniformly on both rods. Determine the magnitude and direction of the electric field E at the center of the circle, proceeding in the steps given below.

(a) Use a superposition argument to transform this problem into a simpler one that involves only one semi-circle of charge. Circle the case shown below that gives the same physical situation as the original problem.

Briefly justify your choice:

The +q semi-circle in quadrant I and II, has same effect as a –q semi-circle in quadrant III and IV. Thus, -q added to –2q semi-circle results in net –3q.

(b) Hopefully, you figured out that the "-3 q" semi-circle was the correct one! Choose an appropriate differential element of charge (call it "dq") on the arc and write down an expression for the magnitude of the electric field |dE| from this element.

l = -3q/pR

Show your argument using a sketch and some brief sentences.

Look at symmetric elements ds and ds’ and draw the resulting differential vectors dE for those elements. Breaking the differential vectors dE into x- and y- axis components, we see that the dEx of the two elements cancel and the dEy components add and are directed downward.

(d) Finally, integrate the expression that you obtained above in (b) and determine the MAGNITUDE of the total electric field. Caution: Remember that your integral should add appropriate COMPONENTS not the total magnitudes of the field from each differential element!

Ignore the signs and put in proper sign of expression at the end.

ds = R dq