query 23

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course Phy 202

8/5 around 2:20AM

023.

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Question: `qIn your own words explain how the introductory experience with

scotch tape illustrates the existence of two types of charge.

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Your Solution:

The experiment shows that each piece of tape has a charge. You can see them

attract/repel each other.

confidence rating #$&*:

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Question: `qIn your own words explain how the introductory experience with

scotch tape supports the idea that the force between two charged particles acts

along a straight line through those particles, either attracting the forces

along this line or repelling along this line.

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Your Solution:

The tape pices move in a line toward each other showing the attraction of

particles in a line from one piece accross to the other.

confidence rating #$&*:

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Question: `qIn your own words explain why this experience doesn't really prove

anything about actual point charges, since neither piece of tape is confined to

a point.

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Your Solution:

The pices of tape have several poin charges that are hard to tell apart. The

different point charges allow the tape freedom of motion.

confidence rating #$&*:

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Question: `qIf one piece of tape is centered at point A and the other at point

B, then let AB_v stand for the vector whose initial point is A and whose

terminal point is B, and let AB_u stand for a vector of magnitude 1 whose

direction is the same as that of AB_u.

Let BA_v and BA_u stand for the analogous vectors from B to A.

Vectors of length 1 are called unit vectors.

If the pieces attract, then in the direction of which of the two unit vectors is

the tape at point A pushed or pulled?

If the pieces repel, then in the direction of which of the two unit vectors is

the tape at B pushed or pulled?

Explain.

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Your Solution:

Attraction: Tape A will move in the direction of point B. This means vector AB_v

and vector AB_u move from point A to point B. Ab_u is the unit vector because

tape A is pused in the direction of AB_u at a magnitude of length 1.

Repulsion:

Tape B repulses tape A moving from point A to point B. It moves the same

direction as AB_u

confidence rating #$&*:

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Given Solution:

If the pieces attract, then the tape at point A is pulled toward point B.

The vectors AB_v and AB_u point from A to B.

Of these the vector AB_u is the unit vector.

So the tape at A experiences a force in the direction of the vector AB_u.

If the pieces repel, then the tape at point B is pushed away from point A.

The direction of the force is therefore from A towards B.

The direction is therefore that of the vector AB_u.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qUsing the notation of the preceding question, which you should have

noted on paper (keep brief running notes as you do qa's and queries so you can

answer 'followup questions' like this), how does the magnitude of the vector

AB_v depend on the magnitude of BA_v, and how does the magnitude of each vector

compare with the distance between A and B?

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Your Solution:

confidence rating #$&*:

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Given Solution:

The distance from A to B is the same as the distance from B to A. So the

vectors AB_v and BA_v have the same length.

Each vector therefore has magnitude equal to the distance between A and B.

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Self-critique (if necessary):

AB v = -BA v = -1 same magnitude just opposite direction

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Self-critique Rating:

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Question: `qUsing the notation of the preceding question, how is the force

experienced by the two pieces of tape influenced by the magnitude of AB_v or

BA_v?

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Your Solution:

AB_u = AB_v / | AB_v |

The AB_v divided by itself will give a unit vector in the direction it is

moving.

confidence rating #$&*:

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Given Solution:

The expected answer is that the force exerted by two charges on one another is

inversely proportional to the square of the distance between them. So as the

magnitudes of the vectors, which are equal to the distance between A and B

(i.e., to the separation between the two pieces of tape), increases the force

decreases with the square of the distance (similarly if the distance decreases

the force increases in the same proportion).

This answer isn't exactly correct, since the two pieces of tape are not point

charges. Since some parts of tape A are closer to B than other parts of tape A,

and vice versa, the inverse square relationship applies only in approximation

for actual pieces of tape.

STUDENT COMMENT: Still not sure how to answer this question. I dont know what

sort of difference there is between the magnitude vector and the unit vector.

INSTRUCTOR RESPONSE: As noted in the preceding solution, AB_v stands for the

vector whose initial point is A and whose terminal point is B, and AB_u stands

for a vector of magnitude 1 whose direction is the same as that of AB_u.

To get a unit vector in the direction of a given vector, you divide that vector

by its magnitude. So AB_u = AB_v / | AB_v |.

The purpose of a unit vector is to represent a direction. If you multiply a unit

vector by a number, you get a vector in the same direction as the unit vector,

whose magnitude is the number by which you multiplied it.

The problem does not at this point ask you to actually calculate these vectors.

However, as an example:

Suppose point A is (4, 5) and point B is (7, 3). Then AB_v is the vector whose

initial point is A and whose terminal point is B, so that AB_v = <3, -2>, the

vector with x component 3 and y component -2.

The magnitude of this vector is sqrt( 3^2 + (-2)^2 ) = sqrt(11).

Therefore AB_u = <3, -2> / sqrt(11) = (3 / sqrt(11), -2 / sqrt(11) ). That is,

AB_u is the vector with x component 3 / sqrt(11) and y component -2 / sqrt(11).

(Note that these components should be written with rationalized denominators as

3 sqrt(11) / 11 and -2 sqrt(11) / 11, so that AB_u = <3 sqrt(11) / 11, -2 sqrt

(11) / 11 > ). The vector AB_u is a unit vector: it has magnitude 1.

The unit vector has the same direction as AB_v. If we want a vector of

magnitude, say, 20 in the direction of this vector, we simply multiply the unit

vector AB_u by 20 (we would obtain 20 * <3 sqrt(11) / 11, -2 sqrt(11) / 11 > =

<60 sqrt(11) / 11, -40 sqrt(11) / 11 >).

You don't need to understand this example at this point, but if you wish to

understand it you should probably sketch this situation and identify all these

quantities in your sketch.

STUDENT RESPONSE

AB_v/(AB_v)absolute value

So AB_u would either be 1 or -1.

INSTRUCTOR COMMENT

AB_u and BA_v are vectors, so they have both magnitude and direction.

Your answer is correct if everything is in one dimension (e.g., if all charges

are on the x axis and all forces directed along the x axis).

In one dimension direction can be specified by + and - signs.

In two dimensions + and - signs are not sufficient. In two dimensions the

direction of a vector is generally specified by its angle as measured

counterclockwise from the positive x axis.

In two dimensions, the magnitude of AB_u would be the same as that of BA_u, but

the vectors would be in the opposite direction.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qQuery introductory set #1, 1-5

Explain how we calculate the magnitude and direction of the electrostatic force

on a given charge at a given point of the x-y plane point due to a given point

charge at the origin.

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Your solution:

use the equation F/Q = kq1/r^2 to find the magniturde and direction of the

electric field. If q1 is positive the electric field will emit a force in a

positive direction. If q1 is negative the electric field will emit a force in a

negatime direction. The magnitude of the force vector = arctan(y/x). 180 degrees

must be added to theta if x is negative.

confidence rating #$&*:

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Given Solution:

`a** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where

q1 is the charge at the origin. The force is one of repulsion if q1 and Q are of

like sign, attraction if the signs are unlike.

The force is therefore directly away from the origin (if q1 and Q are of like

sign) or directly toward the origin (if q1 and Q are of unlike sign).

To find the direction of this displacement vector we find arctan(y / x), adding

180 deg if x is negative. If the q1 and Q are like charges then this is the

direction of the field. If q1 and Q are unlike then the direction of the field

is opposite this direction. The angle of the field would therefore be 180

degrees greater or less than this angle.**

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Self-critique (if necessary):

If q1 is positive the force moves away from the origin, if negative toward the

origin. If q1 is positive then arctan(y/x) is the direction of the electric

field. If q2 is negative the fild is in the opposite direction or arctan(y/x +

180).

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Self-critique Rating:

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Question: `qQuery introductory set #1, 1-5

Explain how we calculate the magnitude and direction of the electrostatic force

on a given charge at a given point of the x-y plane point due to a given point

charge at the origin.

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Your solution:

use the equation F/Q = kq1/r^2 to find the magniturde and direction of the

electric field. If q1 is positive the electric field will emit a force in a

positive direction. If q1 is negative the electric field will emit a force in a

negatime direction. The magnitude of the force vector = arctan(y/x). 180 degrees

must be added to theta if x is negative.

confidence rating #$&*:

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Given Solution:

`a** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where

q1 is the charge at the origin. The force is one of repulsion if q1 and Q are of

like sign, attraction if the signs are unlike.

The force is therefore directly away from the origin (if q1 and Q are of like

sign) or directly toward the origin (if q1 and Q are of unlike sign).

To find the direction of this displacement vector we find arctan(y / x), adding

180 deg if x is negative. If the q1 and Q are like charges then this is the

direction of the field. If q1 and Q are unlike then the direction of the field

is opposite this direction. The angle of the field would therefore be 180

degrees greater or less than this angle.**

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Self-critique (if necessary):

If q1 is positive the force moves away from the origin, if negative toward the

origin. If q1 is positive then arctan(y/x) is the direction of the electric

field. If q2 is negative the fild is in the opposite direction or arctan(y/x +

180).

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Self-critique Rating:

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Question: `qQuery introductory set #1, 1-5

Explain how we calculate the magnitude and direction of the electrostatic force

on a given charge at a given point of the x-y plane point due to a given point

charge at the origin.

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Your solution:

use the equation F/Q = kq1/r^2 to find the magniturde and direction of the

electric field. If q1 is positive the electric field will emit a force in a

positive direction. If q1 is negative the electric field will emit a force in a

negatime direction. The magnitude of the force vector = arctan(y/x). 180 degrees

must be added to theta if x is negative.

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`a** The magnitude of the force on a test charge Q is F = k q1 * Q / r^2, where

q1 is the charge at the origin. The force is one of repulsion if q1 and Q are of

like sign, attraction if the signs are unlike.

The force is therefore directly away from the origin (if q1 and Q are of like

sign) or directly toward the origin (if q1 and Q are of unlike sign).

To find the direction of this displacement vector we find arctan(y / x), adding

180 deg if x is negative. If the q1 and Q are like charges then this is the

direction of the field. If q1 and Q are unlike then the direction of the field

is opposite this direction. The angle of the field would therefore be 180

degrees greater or less than this angle.**

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Self-critique (if necessary):

If q1 is positive the force moves away from the origin, if negative toward the

origin. If q1 is positive then arctan(y/x) is the direction of the electric

field. If q2 is negative the fild is in the opposite direction or arctan(y/x +

180).

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Self-critique Rating:

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&#This looks good. Let me know if you have any questions. &#