QUESTION 1 .... Two equal charges of +9 micro-coulombs are initially some distan

Question

QUESTION 1 .... Two equal charges of +9 micro-coulombs are initially some distance apart and experience some magnitude of electric force. They then exchange charge (electrons) and one charge gives some electrons to the other charge somehow. However, after the exchange both charges still remain positive. After the exchange, they are then moved further apart by a factor of 2.5. Also, the magnitude of the electric force is decreased by a factor of 13 for the new charges at the new distance. What is the magnitude (absolute value) of the charge exchanged between the two charges in micro-coulombs?

QUESTION 2... A charge of -7 micro-coulombs lies at the origin. Another charge of +9 micro-coulombs lies at x = +0.66 meters, y = 0 meters, and an unknown charge lies at x = +0.76 meters, y = +0.06 meters.   The angle of the total electric force vector on the charge at the origin is 224 degrees counter-clockwise from the positive x axis. What is the value of the unknown charge in micro-coulombs? If the answer is negative, include a negative sign, but do not include a positive sign for positive answers.

Explanation / Answer

Question 1

initial    configuration

. q1          q2

           r

F12

. q1                          q2      end   cofiguration

          2.5r

F´12=   F12/13

F12 = k q1q2/r2

F´12 = k q´1 q´2/ r´2        F12/13 = k q´1 q´2 / ( 2.5r)2

. q1=q2= 9 10-6 C

.   k q1 q2/ (r2 13) = k q´1 q´2 / 6.25 r2

.   q1 q2/ 13 = q´1 q´2/6.25          9 10-6   9 10-6= 13/6.25   q´1 q´2

81 10-12 = 2.08 q´1 q´2

. q´1 q´2 = 38.94 10-12

The total charge must be constant

.   q1+q2 = q´1+q´2                          18 10-6 = q´1 +q´2

We write the set of equations

. q´1 q´2 = 38.94 10-12

. q´1 +q´2 = 18 10-6

We resolved the system

. q´1 = 38.94 10-12 / q´2

.   q´1 = 18 10-6 - q´2

. 18 10-6 - q´2 =38.94 10-12 / q´2                                 

.      18 10-6 q´2 - q´22 =38.94 10-12

.      q´22 - 18   10-6 q´2 + 38 10-12 =0

     .     q´2 = ( 18 10-6 ±Ö( 324 10-12 – 4 1   38 10-12) )/2

.    q´2 =    (18 10-6 ± 13.12 10-6)/2

. q´2 =   15.56    10-6 C           

                  2.44   10-6 C

. q´1 = 18 10-6 - q´2

. q´1 =   18 10-6 - 15.56 = 2.44    10-6 C

. q´1 = 18 10-6 - 2.44     = 15.56    10-6 C

question 2

We use Newton's second law and the law of Coulomb

F12 = k q1 q2 /r122    i              

Data

. q1= -7 10-6 C

. q2= + 9 10-6 C

. r12 = 0.66 m

F12 = 9 109 7 10-6 9 10-6 /0.662 = 1301.65 10-3 N             F12= 1.301 N

F13 = 9 109   7 10-6 q3 / ( 0.762 + 0.062) = 108.32 103 q3 N

We are looking for the resulting force on each axis

Ft     angle = 224°     224-180= 44°     below the x axis

Ftx = - FT Cos 44                         Fty = - FT Sin 44

We are looking for the components of F13

Tg angle = 0.06/0.76      angle = tg-1( 0.078)     angle = 4.5°  

F13x = F13 Cos 4.5                     F13y=   F13 Sin 4.5

We calculate the total force

Ftx= F12 –F13x               -Ft Cos 44= 1.301 – 108.32 103 q3 Cos (4.5)               

Fty = -F13y                      -Ft Sin 44 = - 108.32 103 q3 Sin (4.5)

.    – (108.32 103 q3 Sin (4.5)/Sin44 ) Cos 44= 1.301 – 108.32 103 q3 Cos (4.5)               

.    ( 108.30 103 Sin 4.5    cTg 44)   q3 = -1.301 + 108.32 103 cos 4.5   q3

. q3 108.32 103 (cos 4.5 +Sin 4.5 cTg 44) = -1.301

. q3 = -1.301/(108.32 103 (cos 4.5 +Sin 4.5 cTg 44) = -1.301 10-3 / 108.32( 0.9969+0.07846 1.0355)

.q3 = - 1.301 10-3 /116.78

. q3 = -11.14 10-6 C

Q3 is a negative charge

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