Coulomb's Law Lab

Name: Section: SP1L-17 Date: 2/26/11

Coulomb’s Law: What is the charge on a charged pith ball?

Experimental Determination of the Electrostatic Force acting between two charged pith balls

Equipment: a pvc pipe and fur or wool, a metric ruler, a protractor, two pith balls on a stand as in figure 1. Your teacher will tell you the mass of your pith balls.

Procedure: Make sure the strings of your two pith balls are untangled as in figure 1. Make sure that the pith balls are neutral by rubbing them with your finger. Measure the length from the point that the strings can pivot on the stand to the center of the ball and record it. Charge the pith balls according to the instructions given by your lab teacher. When the pith balls are
fully charged alike (negative) they will noticeable repel each other.

figure 1 figure 2

[pic] [pic]

Use the protractor to measure the angle between the two strings.

|Length of the strings, (l) |Angle between the strings, (θ) |
|0.18 m |8 degrees |

When the pith balls have reached their final charged position and are at rest, what do we know about the forces acting on them?

Name the force: A B C

To determine the electrostatic force, label the vector diagram below of the forces acting on the pith ball.

1. Using your vector diagram above, determine the ratio of Fe to Fg in terms of θ/2.

2. Write an equation for the electrostatic force on the pith ball in terms of angle Ѳ/2 and the weight of the pith ball.

3. Calculate r/2 (half the distance between the two charged pith balls) using the length (l) and the angle Ѳ/2. Then double it to find r. Show your work.

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.

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.

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.

2. What if the angle between the pith balls is 40˚ and the mass of each pith ball had a mass of .3 kg. The length of the string is .3 m. Find

a. The weight of the pith ball.

b. the electrostatic force between them
and

c.
the charge on each ball, assuming their charges are equal.

-----------------------
C

B

A

The sum of the forces acting on them is 0.

Electrostatic

gravity

tension

Ѳ/2

Fg

Ft

Fe

Fe/Fg = tan θ/2

Fe= tan θ/2 (Fg)

Tan 4 = r/2/0.18 m tan 4(0.18 m)= r/2 (2) tan 4(0.18 m)= r

r= 0.0252 m

tan θ/2 (Fg)= k (q) (q) / r 2 (tan θ/2 (Fg)r2 )/ k = q2 ((tan θ/2 (Fg)r2)/ k)1/2 = q

(tan θ/2 (Fg)r2 )/ k = q2 (tan 4)(9.81 m/s2)(0.252 m2)/ 8.99 x 109 = q2

q2= 4.846 x 10-12 q = 2.2 x 10-6 C

The sources of error could have been in measuring the angle o⁲敬杮桴漠⁦桴⁥瑳楲杮‮桔⁥瑳楲杮洠杩瑨栠癡⁥敢湥愠氠瑩汴⁥桳牯整r length of the string. The string might have been a little shorter or longer than measured. The angle could have varied based on the electrostatic they already had between them at the moment. The angle could have been slight smaller or larger than the measurement taken by the protractor.

F=ma (0.3 kg)(9.81 m/s/s)= 2.943 N

Fe= tan 20(2.943 N)= 1.071 N/C

Tan 20 = r/2/0.3 m tan 20(0.3 m)(2)= 1.343 m = r Fe = k (q1) (q2) / r 2

6.584 = 8.99 x 109 (q2)/ 1.343 m2 6.584 (1.343 m2) /8.99 x 109 =q2

6.584 (1.804) /8.99 x 109 = 11.878/8.99 x 109 = 1.321 x 10-9=q2

q= 3.635 x 10-5
C