Electrostatic Part 2 – Pith Ball Lab Purpose: The purpose of this lab is to be able to determine the electrostatic force between two charged pith balls by using the mass, the length of the strings and the angle between the strings of the two pith balls.
Apparatus: Figure 1
Figure 2
Procedure: 1.
Obtain Obtain the equipment equipment for the lab: the pvc pipe, fur or wool, ruler, ruler, protractor, protractor, and two pith balls on a stand. Untangl Untanglee the string stringss of the two pith pith balls balls on the stand. stand. Rub both both pith pith balls balls with with your your finger finger to neutral neutralize ize them. them. Using a ruler, ruler, record record the length length of the string string from the focal point to the center of the ball. Charge the pipe with the fur/wool, fur/wool, and place the pipe near the pith balls. balls. When both pith balls repel each other, measure measure the angle at which both of the strings form. Record Record the data, data, and answer answer the follow following ing questi questions ons..
2. 3. 4. 5. 6. 7.
Data: Leng Le ngth th of the the stri string ngss (l) (l) 16 cm = 0.16 m
Angl Anglee betw betwee een n the the stri string ngss (θ) (θ) 8°
Conclusion: a) The electrostatic force produced by the two pith balls can be calculated using
Coulomb's electric force law. The electrostatic force (Fe) was found by multiplying the electrostatic constant (k = 8.99 x 109) by charges of the two pith balls (q1 and q2). Then we divide this product by the square of length of the distance between the two pith balls. b) As the distance between the pith balls increase, the electrostatic force decreases whereas when the distance decrease, the electrostatic force increases. c) From this experiment, we found out that the charges of the pith balls could have been inaccurate if we measured the angle incorrectly with the protractor. The pith balls could have changed charges during the experiment due to their nature of being easily affected by charge.
Lab Questions: When the pith balls have reached their final charged position and are at rest, what do we know about the forces acting on them? The pith balls both repel each other as they are both oppositely charged.
T (Tension) Name the force: A
Fe (Force Electricity)
B
C
Fg (Force Gravity)
To determine the electrostatic force, label the vector diagram below of the forces acting θ /2 on the pith ball.
Fg
T
Fe
1. Using your vector diagram above, determine the ratio of Fe to Fg in terms of θ/2.
tan(θ /2) = Fe/Fg 2. Write an equation for the electrostatic force on the pith ball in terms of angle θ /2
and the weight of the pith ball. Fe = Fg*tan(θ /2)
3. Calculate r/2 (half the distance between the two charged pith balls) using the
length (l) and the angle θ /2. Then double it to find r. Show your work. r 2 r
=
length * sin( θ / 2) 2
= 0.16sin4/(2)
2 r = 0.01m 2
r = length*sin(θ /2) r = 16sin(4) r = 0.02m
4. Set the equation for Fe written in 2. equal to Coulombs Electrostatic Force Law
Fe = k (q1) (q2) / r 2 Assume q1 and q2 are equal and replace them both with q. Solve for q. mg*tan(θ /2) = mg*tan(θ /2) =
q=
k * q1 * q2 r 2 k * q 2 r 2
r 2 * F g * tan(
/ 2) θ
k
5. To calculate the charge q, assume that both charges are equal, and remember to
use the distance measured between the charged pith balls when solving Coulombs’ Law to determine the charge on each pith ball. Fe = q= q=
k * q
2
r 2 r 2 * F g * tan(
θ / 2)
k 5 4 (0.07 )(1*10 − kg )( 4 *10 − )
8.99 *10 9
q = 3.12*10-20 C
Analysis 1. What sources of error might have contributed to the accuracy of your final calculation for q?
Justify your answer with reference to the possible erroneous and inexact measurements used. Error in measurement of the angle and length contribute to the accuracy of the final measurements for q. Human error in measurement is present in this experiment. 2. What if the angle between the pith balls is 40° and the mass of each pith ball
had a mass of 0.3 kg. The length of the string is 0.3 m. Find a. The weight of the pith ball. weight = mg mg = 0.3kg x 9.8m/s2 = 2.94 kg m/s2 b. the electrostatic force between them and tan 20° = Fe/weight 0.364 = Fe/2.94 N Fe = 1.07 N c. the charge on each ball, assuming their charges are equal. q=
