A negatively ncharged particle of lmass 5.7 x 10^-19 kg is moving with a speed o

Question

A negatively ncharged particle of lmass 5.7 x 10^-19 kg is moving with a speed of 35.0 m/s when it enters the region between two parallel capacitor plates. the velocity of the charge is parallel to the plate surfaces and in the positive x-direction. the plates are square with a side of 1.0 cm and the voltage across the plates is 3.35 V. If the particle is initially 1.00 mm from both plates, and it just barely clears the positive plate after traveling 1.00 cm through the region between the plates, how many excess electrons are on the particle? You may neglect gravitational and edge effects. The distance between the plates is 2.00 mm.

Explanation / Answer

Distance between the two parallel plates is d    = 2.00 mm                                                                        = 2.00 mm ( 10-3 m / 1mm )                                                                        =   2.0 *10-3 m                                                                        = 2.00 mm ( 10-3 m / 1mm )                                                                        =   2.0 *10-3 m The potential differece between the two plates is V = 3.35 V The intial speed of the particle is along x -axis is vx = 35 m/s The charge of the electron is e = 1.6*10-19 C The vetricle distance between the plates and the particle is   y = 1.00 mm                                                                                                 = 1.00 mm ( 10-3 m / 1mm )                                                                                                =   1.0 *10-3 m Mass of the particle is m = 5.7*10-19 kg Horizontal distance x = 1.0 cm                                     = 1.0 cm ( 10-2 m /1 cm)                                     = 1.0*10-2 m The time interval for this particle is between the parallel plates is                                            t = x / vx                              ...... (1) The distance travelled by the particle in y direction is                                            y = ( 1/2) a ( t )2 On substituing equation (1) in the wbove equation we get                                          y = ( 1/2) a ( x / vx    )2 The accelartaion of the particle is                                           a = 2 (  y ) (vx )2 / x             ...... (2) ----------------------------------------------------------------------- Accoerding to Newtons second law of motion                                           F = ma The electtic force acting on the electron is                                           F = NeE Compariring the above two equations, we get                                        ma =NeE                                           a = NeE / m          Electric field                        E = V / d                       ...... (3) Here , N = number of excess electrons                                           From equations (2) and (3), we get              2 (  y ) (vx )2 / x = Ne( V / d  ) / m    Rearraging the above equation , we get Number of exess electrons is                                             N = 2 d*m ( y ) (vx )2 /   ( eV *x ) Substitting all values in the above equation, we get   N = 2 (  2.0 *10-3 m )( 5.7*10-19 kg ) (  1.0 *10-3 m ) ( 35 m/s ) / (  (3.35 V  ) (1.6*10-19 C  )(0.01 m) 2                                                   N = 52 electrons                                                                                                    

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