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 …show more content…
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 …show more content…
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
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 …show more content…
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 …show more content…
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