Directions: Questions 1, 2, and 3 are short free-response...

Directions: Questions 1, 2, and 3 are short free-response questions that require about 13 minutes to answer and are worth 8 points. Questions 4 and 5 are long free-response questions that require about 25 minutes each to answer and are worth 13 points each. Show your work for each part in the space provided after that part. (Diagram is not available to copy) In the above diagram, two small objects, each with a charge of –4.0 nC, are held together by a 0.020 m length of insulating string. The objects are initially at rest on a horizontal, non-conducting, friction less surface. The effects of gravity on each other can be considered negligible. (a) Calculate the tension in the string. (b) Illustrate the electric field by drawing electric field lines for the two objects on the following diagram. (Diagram is not available to copy) The masses of the objects are $m_{1}=0.030 \mathrm{kg}$ and $m_{2}=0.060 \mathrm{kg}$ . The string is now cut. (c) Calculate the magnitude of the initial acceleration of each object. (d) On the axes below, sketch a graph of the acceleration $a$ of the object of mass $m_{2}$ versus the distance $d$ between the objects after the string has been cut. (Graph is not available to copy) (e) In a brief paragraph, describe the speed of the objects as time increases, assuming that the objects remain on the horizontal, non-conducting friction less surface.

Directions: Questions 1, 2, and 3 are short free-response questions that require about 13 minutes to answer and are worth 8 points. Questions 4 and 5 are long free-response questions that require about 25 minutes each to answer and are worth 13 points each. Show your work for each part in the space provided after that part.

(Diagram is not available to copy)

In the above diagram, two small objects, each with a charge of –4.0 nC, are held together by a 0.020 m length of insulating string. The objects are initially at rest on a horizontal, non-conducting, friction less surface. The effects of gravity on each other can be considered negligible.
(a) Calculate the tension in the string.
(b) Illustrate the electric field by drawing electric field lines for the two objects on the following diagram.

(Diagram is not available to copy)

The masses of the objects are $m_{1}=0.030 \mathrm{kg}$ and $m_{2}=0.060 \mathrm{kg}$ . The string is now cut.
(c) Calculate the magnitude of the initial acceleration of each object.
(d) On the axes below, sketch a graph of the acceleration $a$ of the object of mass $m_{2}$ versus the distance $d$ between the objects after the string has been cut.

(Graph is not available to copy)

(e) In a brief paragraph, describe the speed of the objects as time increases, assuming that the objects remain on the horizontal, non-conducting friction less surface.

Key Concepts

Graphical Analysis of Force Dependencies

Graphical analysis in physics involves plotting variables such as acceleration against distance to reveal the underlying mathematical relationships, such as those described by inverse-square laws. This skill is crucial for interpreting how changes in separation between objects affect the force, and therefore the acceleration, acting on them.

Electric Field and Field Lines

An electric field represents the forces that a charge would experience per unit charge at any point in space due to other charges. Field lines provide a graphical representation of this field, indicating both the magnitude and the direction of the force. This concept is essential for visualizing how charged objects influence the space around them.

Tension in a Connecting String

Tension is the force transmitted through a string when forces act at its ends, ensuring equilibrium between connected objects. In the context of charged objects held together by a string, calculating the tension involves balancing the electrostatic repulsive force with the mechanical force provided by the string, prior to any disturbances such as cutting the string.

Newton's Second Law

Newton's Second Law relates the net force acting on an object to its mass and the acceleration it experiences (F = ma). Applying this principle allows us to predict the motion of charged objects once the forces, such as those from electrostatic interactions, are known, especially in situations where additional constraints are removed.

Coulomb's Law

Coulomb's Law quantifies the electrostatic force between two point charges; it states that the force is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This law is fundamental in understanding the interactions between charged objects and is used to calculate the magnitude and direction of electric forces in static systems.

Transcript

00:02 In this example, we are going to look at forces and electric field of point charges and see how the force on point charges affects their motion.

00:17 So the situation that we have are two equally charged point charges.

00:26 We'll treat them like point charges, and they're tied together with a lightweight string.

00:31 Both of them have a charge of minus 4 nanoculums, which is times 10 to the minus 9 quulums.

00:49 We'll call this one charge 1 and charge 2.

00:54 So they're held together by the string, and what we know is that there's a repulsion that acts equally and oppositely on both of these charges.

01:07 So the force of repulsion is acting on both of those, and that's going to create a tension in the string that's equal and opposite as long as that string is lightweight.

01:27 So each of those charges is going to be held in equilibrium by that interactive tension.

01:36 Okay.

01:38 And what we see from this diagram is the tension has to be equal in magnitude to the cholome force of repulsion between the two charges.

01:50 So that's where our electrostatics equation is going to come into play.

01:55 Kulam's law says that the force between two charges is k times the absolute magnitude of the two charges interacting over their separation squared.

02:13 Here we're given that the separation is 2 centimeters.

02:17 So 0 .02 meters is r d.

02:28 And we can work this out, kulom's law, simply by using the values we have in the electrical constant k in the appropriate units.

02:46 Si is 9 times 10 to the 9th.

02:52 Charges have equal magnitudes 4 times 10 to the minus 9th.

02:58 We'll square that and divide by the distance squared.

03:04 And that will come out to be in newton's.

03:09 Working that out, we see this force is, 3 .6 times 10 to the minus 4th newton's.

03:25 This is also equal to the tension that's pulling back on, equally on both sides of the string.

03:37 And we may want to show the electric field in the vicinity around those two charges.

03:45 Let's go ahead and do that, two negative charges, equal signs, equal magnitudes.

03:53 But basically, the electric field shows the lines of force that a positive charge would experience.

04:02 So let me draw those in blue.

04:07 And what i'm thinking of is the total electric field coming from both of those charges.

04:13 If you're near one of the charges on the side opposite, the other charge, the electric field will simply point straight into each charge.

04:24 Charge representing the attraction of a positive charge.

04:29 You'd have a fairly strong, straight electric field on that side.

04:35 As you get into the middle, though, a positive charge is going to have a hard time figuring out whether it's going to attract to the left charge or the right charge.

04:48 And those electric fields are actually going to partially cancel in that space in between.

04:56 And so you get kind of this saddle.

05:01 A saddle always kind of represents an ambiguity in direction.

05:12 And i really should be showing those electric field lines really spaced out in the area, in between the two charges, representing a weak electric field in that region...

Please give Ace some feedback