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A small object with mass m, charge q, and initial speed v0 = 4.00×103 m/s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm (Figure 1) . The electric field between the plates is directed downward and has magnitude E = 600 N/C . Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distance d = 1.35 cm from its original direction of motion and reaches a collecting plate that is 56.0 cm from the edge of the parallel plates. Ignore gravity and air resistance.Calculate the object's charge-to-mass ratio, q/m.

 Apr 10, 2016
 #1
avatar+19687 
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I'll get back to this shortly, but I think the key is to recognize the 'x' velocity does not change.....the acceleration of the particle is the velocity imparted in the 'y' direction and it is 1.35/56 proportional to the initial 'x' velocity of  4 x 103 m/s    then you can use : (Lorentz force law) and  Newtons second law to calculate B    then    m/q  a  =  E + v x B         More later.....gotta run....

 Apr 11, 2016
 #2
avatar+19687 
0

OK.....the y velocity is .0135/t    where t is the time it took to travel 56 cm at   (4 x 103m/s)  = .56/412

.0135/ (.56/412) = ( .0135 x 412) /.56 = 9.932 m/s

This   velocity change from ZERO to this value occured in the time it took to cover the plate length of 26cm (.26m) at 4 x 103 m/s.

(is this written properly?  4 x 103 m/s?)        412m/s (pretty slow)    .26/412 = .00063107 seconds

 

Acceleration in the y direction is thus the change in

velocity/time = (9.932)/.00063107 = 15738.4 m/s^2

 

Since F= ma    Thus the force is    15738.4 ( m )

Lorentz force law :  F= q(E+ v x B) = 15738.4 m    divide both sides by m

q/m (E + v x B) = 15738.4

or rewritten:

q/m = 15738.4/ (E + v x B)   We are given E  and v which we can substitute....but we still do not have 'B'   the magnetic flux density......hmmmmm   .... NOW I'M STUCK!    Wonder if I was given all of the information?   And was the initial speed really  4 x 103 m/s ?

Think I might be on the wrong path???

 Apr 11, 2016
 #3
avatar+19687 
0

(Re-posted with v= 4 x 10^3   NOT  NOT  4 x 103 m/s)

 

OK.....the y velocity is .0135/t    where t is the time it took to travel 56 cm at   (4 x 10^3m/s)  = .56/4000

.0135/ (.56/4000) = ( .0135 x 4000) /.56 = 96.429 m/s

This   velocity change from ZERO to this value occured in the time it took to cover the plate length of 26cm (.26m) at 4 x 10^3 m/s.

   .26/4000 = 6.5 x10^5 seconds

 

Acceleration in the y direction is thus the change in

velocity/time = (96.429)/.6.5 x 10^-5 = 1483523  m/s^2

 

Since F= ma    Thus the force is   1483523 ( m )

Lorentz force law :  F= q(E+ v x B) = 1483523 m    divide both sides by m

q/m (E + v x B) = 1483523

or rewritten:

q/m = 1483523/ (E + v x B)   We are given E  and v which we can substitute....but we still do not have 'B'   the magnetic flux density......hmmmmm   .... NOW I'M STUCK!    Wonder if I was given all of the information? 

Think I might be on the wrong path???

 Apr 11, 2016

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