You are working in cooperation with the Public Health department to design an el

Question

You are working in cooperation with the Public Health department to design an electrostatic trap for particles from auto emissions. In order to verify the correct operation of the trap the design engineers must be able to monitor the speed of the particles before they are trapped. If the particles are moving too fast then they cannot be effectively trapped. The particular detector the engineers want to use is only sensitive to particles moving less than 1000 m/s.The following design parameters are given.

The average emission particle that enters the device is ionized by exposure to ultraviolet radiation resulting in the removal of electrons so that it has a charge of +3.0 x 10-8 C.
The trapping element is a single large square plate negatively charged with a charge density of -8.0 x 10-6 C/m2.
The average particle in the emission stream is moving 900 m/s when it is 15 cm from the plate.
The detector for the particles is located 7.0 cm from the plate.
An average emission particle has a mass of 6.0 x 10-9 kg.

Does this electrostatic trap meets the design parameters?

Explanation / Answer

Given charge on the par ticle   q =   3*10-8 C mass of the particle   m =   6*10 -9 kg velocity of the plate at x = 15 cm is   900 m/s surface charge density of square plate   = - 8 *10 -6 C / m 2   electric field due to square plate is                          E = / 2 o                                 = ( 8*10 -6 ) / 2 ( 8.85* 10 -12 )                               = 4.51977*105 V / m elctric force on the particle due to square plate is                          F = E q                              = ( 4.51977*105 V / m ) ( 3*10-8 C )                              = 1.3559*10-2 N    acceleration of the particle is                             a = F / m                               = 1.3559*10-2 N   / 6*10 -9 kg                                 = 2.25988 * 10 6 m / s 2         velocity of the particle at x =   7 m from the plate is                           ( 1/2) m v 2 = V q                                               = E / d * q                                        v 2   = 2 Eq / d m                                        v 2 = (4.51977*105 V / m ) 2 (3*10-8 C ) / 0.07 * 6*10 -9 kg                                           v    = 8.035*103    m/s      here the charge between particle and plate opposite.so the force is attractive force        so velocity of the particle is graduallyy increases                    particle reaches the detecor but the dector is not detected because the velocity of the particle at the detector is grater than 1000 m/s                              velocity of the particle at x =   7 m from the plate is                           ( 1/2) m v 2 = V q                                               = E / d * q                                        v 2   = 2 Eq / d m                                        v 2 = (4.51977*105 V / m ) 2 (3*10-8 C ) / 0.07 * 6*10 -9 kg                                           v    = 8.035*103    m/s      here the charge between particle and plate opposite.so the force is attractive force        so velocity of the particle is graduallyy increases                    particle reaches the detecor but the dector is not detected because the velocity of the particle at the detector is grater than 1000 m/s                        

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