Three particles are arranged on coordinate axes as shown above. Particle A has a

Question

Three particles are arranged on coordinate axes as shown above. Particle A has a charge Q(a)= -.2nC, and is initially on the y-axis at y=.03m The other two particles each have charges Q(b)= +.3nC and are held faxed on the x-axis= -.04m and x= +.04m, respectively.

a) calculate the magnitude of the net electric force on particle A when it is at y=.03m and state its direction.

b) particle A is ten realeased from rest. Qualitatively describe its motion over a long time. detailed please.

c) calculate the magnitude of the force the magnetic field exerts on particle A as it enters the magnetic field

d) An electric field can be applied to keep particle A moving in a straight line through the magnetic field. Calculate the magnitude of this electric field and state its direction

Explanation / Answer

Given charges   Q (a ) = -0.2 nC                Q ( b) = 0.3 nC                 Q (c ) = 0.3 nC distance between charge Q(b) and Q (a) is              ra = rb = r = 0.032 + 0.042                                 = 0.05 m    angle   = tan -1   ( 0.03 / 0.04 )                   = 36.8689 o   force on Q(a)   due to charges    Q (b) and Q (c) is           Fb = Fc    =   k Q(a) Q(b) / r 2                                 = ( 9*109 ) ( 0.2*10-9 ) ( 0.3*10-9 ) / 0.05 2                                 = 216 * 10 -9 N componenets of force is          Fbx   = Fb cos                 = ( 216 * 10 -9 N ) cos 36.8689                  = 172 * 10-9         Fby = - Fb sin                  = ( 216 * 10 -9 N ) sin 36.8689                  = -129.59 * 10-9          Fcx = - Fc cos                = -172 * 10-9        F cy = - Fc sin                 = -   -129.59 * 10-9   componenets of net force on Q(a) due to Q (b) and Q (c) is              Fx = Fbx + Fcx                     = 0               Fy = - 2 (  -129.59 * 10-9 )                     = - 259.1939 * 10 -9       magnitude of net force is            F = 259.1939 * 10 -9      N   direction is           = tan -1 (- 259.1939 * 10 -9      N / 0 )              = - 90 degrees    ______________________________________________________ b)    charge on particle A is   Q(a) i s q = 0.2 nC      magnitue of magnetic field is B = 0.5 T    speed of particle is   v = 6000 m / s     magnetic force on particle                  F = B q   v                     = 0.5 * 0.2*10-9 * 6000                      = 0.6 N   c)      Force on charged particle due to magnetic field and electric field is             F = B q v + E q                 = q ( B v + E )     plug the values anf get the value of electric field                  = -129.59 * 10-9          Fcx = - Fc cos                = -172 * 10-9        F cy = - Fc sin                 = -   -129.59 * 10-9   componenets of net force on Q(a) due to Q (b) and Q (c) is              Fx = Fbx + Fcx                     = 0               Fy = - 2 (  -129.59 * 10-9 )                     = - 259.1939 * 10 -9       magnitude of net force is            F = 259.1939 * 10 -9      N   direction is           = tan -1 (- 259.1939 * 10 -9      N / 0 )              = - 90 degrees    ______________________________________________________ b)    charge on particle A is   Q(a) i s q = 0.2 nC      magnitue of magnetic field is B = 0.5 T    speed of particle is   v = 6000 m / s     magnetic force on particle                  F = B q   v                     = 0.5 * 0.2*10-9 * 6000                      = 0.6 N   c)      Force on charged particle due to magnetic field and electric field is             F = B q v + E q                 = q ( B v + E )     plug the values anf get the value of electric field   

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