Coulomb\'s Law Activity Sheet PDF

Preview

Summary

Download Coulomb's Law Activity Sheet PDF

Description

Coulomb’s Law Virtual Laboratory Experimentation is the greatest science. - Arabic Proverb Coulomb’s Law describes the electrostatic force created by charged bodies. According to Coulomb’s Law, the force of attraction (or repulsion) between two charged bodies is directly proportional to the product of their charged magnitudes and inversely proportional to the square of the distance between them. In equation,

F =k

q1q2 r2

where: k = Coulomb’s Constant = 9.0x109 Nm2/C2 q1 = charge on mass 1 q2 = charge on mass 2 r = the distance between the two charges In this part of the module, you will explore the Coulomb’s Law Interactive Simulation to investigate the relationship among electrostatic force, charge magnitudes and the distance separation between charges. Learning Objectives: 1. To determine how the magnitude of electrostatic force changes as a function of charge magnitudes and distance between charged objects 2. To compare the electrostatic force exerted by charged bodies on each other 3. Verify the results of the experiment using Coulomb’s Law Materials: Simulation Title: Coulomb’s Law https://phet.colorado.edu/sims/html/coulombs-law/latest/coulombs-law_en.html Procedure: 1. Once the program is loaded, select the Macro Scale option and familiarize yourself with the components of the simulation.

2. The magnitude of the two charged bodies can be adjusted by clicking the button or moving the glider. 3. The distance between charged bodies can be increased or decreased by moving the charges left and right. Use the given ruler to measure the distance. The separation distance, r, between charged bodies is taken as the distance between their centers. 4. Explore the Coulomb’s Law Interactive Simulation by completing the table and answering the guide questions. --------------------------------------------------------------------------------------------------------------------A. Effect of Varying Distance to Electrostatic Force 1. Determine how the separation distance between charged objects affects electrostatic force between them. Assign the charge of q1 and q2 and hold it constant. Let the initial distance between charges be 4.0cm. Charge 1, q1 =

-2

.

Charge 2, q2 =

6

.

Data Table1: Effect of Varying Distance to Electrostatic Force Separation distance, r, Electrostatic Force, F, Cm (N) 1. Initial distance 4 67.407 2. Double the distance 8 16.852 3. Reduced the distance to 2 269.627 ½

2. Verify the results of the experiment. Compute for the electrostatic force for the three conditions using Coulomb’s Law. Do not forget to convert to standard units – Coulomb (C) for q, meter (m) for r, and Newton (N) for F. Note: Attach the copy of solution here. q2 = 6μC (1x 10-6C) The simulator showed that there is

1. k= 9.0x109 Nm2/C2 q1 = -2μC (-1x 10-6C) = = 2 x 10-6C

q1=-2 and q2=6 are hold constant with

r = 8cm

=

F =k

67.407N electrostatic force when

= 6 x 10-6C

4 cm distance between each other.

= 0.08m

q1q 2 r2

Moreover, the computation showed 67.5N electrostatic force. Although

(2𝑥10−6 𝐶)(6𝑥10−6 𝐶)

= (9.0X109 Nm2/C2) (

(0.04𝑚)2

they are not exactly the same, the

)

= 67.5 N

q2 = 6μC (1x 10-6C) The simulator showed that there is 16.852N electrostatic force when

= 6 x 10-6C

q1 = -2μC (-1x 10-6C) = = 2 x 10-6C

q1=-2 and q2=6 are hold constant with

r = 8cm

=

8 cm distance between each other.

= 0.08m

q1q 2 r2

Moreover, the computation showed 16.875N electrostatic force. Although

(2𝑥10−6 𝐶)(6𝑥10−6 𝐶)

= (9.0X109 Nm2/C2) (

(0.08𝑚)2

they are not exactly the same, the va-

)

= 16.875N

0.020N difference. Therefore, the

q2 = 6μC (1x 10-6C) The simulator showed that there is 269.627N electrostatic force when

= 6 x 10-6C

q1 = -2μC (-1x 10-6C) = = 2 x 10-6C

q1=-2 and q2=6 are hold constant with

r = 8cm

=

2 cm distance between each other.

= 0.08m

Moreover, the computation showed 270N electrostatic force. Although

(2𝑥10−6 𝐶)(6𝑥10−6 𝐶)

= (9.0X109 Nm2/C2) ( 270 N

lues are close to each other with only

result of the experiment is correct.

3. k= 9.0x109 Nm2/C2

qq F = k 122 r

only 0.1N difference. Therefore, the result of the experiment is correct.

2. k= 9.0x109 Nm2/C2

F =k

values are close to each other with

(0.02𝑚)2

)

they are not exactly the same, the values are close to each other with only 0.373N difference. Therefore, the result of the experiment is correct.

B. Effect of Varying Charges to Electrostatic Force 1. Determine how the charge magnitudes affect the electrostatic force. Assign the value for the separation distance between charges and hold it constant. Let the initial charge of q1 and q2 be 2.0µC. Separation distance, r =

6

.

Data Table2: Effect of Varying Charges to Electrostatic Force Charge 1, q1, Charge 2, q2, µC µC 1. Initial charge 2. Double q1 and triple q2 3. Reduced both q1 and q2 to ½

2.0 4.0

2.0 6.0

Electrostatic Force, F, (N) 9.986 59.917

1.0

1.0

2.497

2. Verify the results of the experiment. Compute for the electrostatic force for the three conditions using Coulomb’s Law. Do not forget to convert to standard units – Coulomb (C) for q, meter (m) for r, and Newton (N) for F. Note: Attach the copy of solution here.

q2 = 2μC (1x 10-6C)

1. k= 9.0x109 Nm2/C2 q1 = 2μC (1x 10-6C)

and q2=2 are hold constant with 6 cm

r = 6cm

=

distance

= 0.06m

F =k =

9.986N electrostatic force when q1=2

= 2 x 10-6C

= = 2 x 10-6 C

q1q 2 r2

each

other.

10N electrostatic force. Although they

= 10 N

are not exactly the same, the values are close to each other with only 0.014N difference. Therefore, the result of the experiment is correct.

q2 = 6μC (1x 10-6C)

2. k= 9.0x109 Nm2/C2 q1 = 4μC (1x 10-6C)

q1=4 and q2=6 are hold constant with

r = 6cm

=

The simulator showed that there is 59.917N electrostatic force when

= 6 x 10-6C

= = 4 x 10-6 C

=

between

Moreover, the computation showed

(2𝑥10−6 𝐶)(2𝑥10−6 𝐶) ) (9.0X109 Nm2/C2) ( (0.06𝑚)2

6 cm distance between each other.

= 0.06m

F =k

The simulator showed that there is

Moreover, the computation showed

q1q 2 r2

60N electrostatic force. Although they

(4𝑥10−6 𝐶)(6𝑥10−6 𝐶) ) (9.0X109 Nm2/C2) ( (0.06𝑚)2

= 60 N

are not exactly the same, the values are close to each other with only 0.083N difference. Therefore, the result of the experiment is correct.

q2 = 1μC (1x 10-6C)

3. k= 9.0x109 Nm2/C2

2.497N electrostatic force when q1=1

= 1 x 10-6C

q1 = 1μC (1x 10-6C) = = 1 x 10-6 C

and q2=1 are hold constant with 6 cm

r = 6cm

=

The simulator showed that there is

distance

= 0.06m

between

each

other.

Moreover, the computation showed 2.5N electrostatic force. Although

F =k

q1q 2 r2

they are not exactly the same, the (1𝑥10−6 𝐶)(1𝑥10−6 𝐶)

= (9.0X109 Nm2/C2) ( = 2.5 N

(0.06𝑚)2

values are close to each other with

)

only 0.003N difference. Therefore, the result of the experiment is correct.

Guide Questions: 1. How does the separation distance affect the magnitude of the electrostatic force between charges? The separation distance inversely affects the magnitude of the electrostatic force between charges. This is because when the separation distance increases the magnitude of electrostatic force between charges decreases. On the other hand, when the separation distance decreases the magnitude of electrostatic for between charges increases. This can be observed in Data Table 1 wherein when the separation distance was 8 cm the magnitude of electrostatic force was 16. 652 N. However, as the distance was decreased to 4 cm the magnitude of electrostatic force was increased to 67.407 N. Moreover, when the separation distance was further decreased to 2 cm the magnitude of electrostatic force increased again and became 269.627 N. 2. How do the magnitudes of the charged bodies affect the electrostatic force between them? The magnitude of the charged bodies have direct effect to electrostatic force between them. This indicates that as the magnitude of force increases the electrostatic force increases as well. Moreover, if the magnitude of charged bodies decreases the electrostatic force between them decreases as well. This can be observed in Data Table 2. Notice that with the initial charge, q1=2 µC and q2=2 µC, the electrostatic force that was present were 9.986 N. However, upon doubling q1 and tripling q2, the electrostatic force increased to 59.917 N. On the other hand, when the initial charges, q1 and q2, were reduced to half, the electrostatic force that was present was also reduced. 3. How does the electrostatic force exerted by q1 on q2 compare to the electrostatic force exerted by q2 on q1? The electrostatic force exerted by q1 on q2 and the electrostatic force exerted by q2 on q1 are similar in value. This means that q1 will repel or attract q2 with a given electrostatic force and q2 will also repel or attract q1 with the same electrostatic force CONCLUSION: Columb’s Law explains that electric force acts on a point charge q1 due to the presence of a second point charge q2. Moreover, with the results gathered from this experiment, I therefore conclude that the separation distance of two charges is inversely proportional to the electrostatic force between two charged objects. However, electrostatic force is directly proportional to the magnitude of force of the charged bodies. Furthermore, the electrostatic force exerted by q1 on q2 is similar to the electrostatic force exerted by q2 on q1 as the force that will be used by q1 to repel or attract q2 will be the same amount of electrostatic force that will be used by q2 to repel or attract q1....