Electric Field- answer

1.1)            The space around a charge where its effect can be felt is called electric field. An electric charge placed at any point in it experiences an electric force. Strength or intensity of the electric field at any point is defined as the force acting on a unit positive charge placed at that point. The unit of electric intensity in SI is N/C or V/m

                E=F/q

1.2)     

                                                            

                                       

1.2)            Electric field at P due to Q1:

               

                                                           E₁ = kQ/r²

                                                              = (9x 10⁹ x14x10^-6)/ 1²

                                                              = 126x10³ NC^-1   to the left

            Electric field at P due to Q2:

  

                                                         E₂ = kQ/r²

                                                              = (9x 10⁹ x20x10^-6)/ 2²

                                                              = 45x10³ NC^-1   to the left

                                                      E_net= E₁+ E₂

                                                            = 126x10³ + 45x10³ = 171 x 10³ NC^-1   to the left

2)                                                                                                                                        V=450V

d=0.15m

     Therefore electric field between the plates, E = V/d

                                                                              = 450/0.15

                                                                           E= 3000 V/m

                   Charge of electron, q = e = 1.6x 10^-19 C

              Force of the electron , F = qE

                                                     =  1.6x 10^-19x 3000 = 4.8x 10^-16 N

       Acceleration of the electron,  a= F/m

                                                        = (4.8x 10^-16)/ ( 9.1x 10^-31)

                                                        = 0.5275 x 10¹⁵ ms^-2

        Considering the motion of the electron from the negative plate to the positive plate, u=0 ,

        a=0.5275 x 10¹⁵ ms^-2 ; S=0.15 m

         S= ut+ at²/2

                      0.15=(0.5275x 10¹⁵xt²)/2

         

          t² = (2x0.15)/ (0.5275x 10¹⁵)

          t = [(2x0.15)/ (0.5275x 10¹⁵)]^1/2 = 2.385 x 10^-8 s

     The velocity with which the electron hits the positive plate, v = u + at

                 i.e,  v = 0 + ( 0.5275x 10¹⁵ x 2.385 x 10^-8 )  

                           v = 1.258 x 10⁷ ms^-1