A pair of charged conducting plates produces a uniform field of E = 5000 N/C, di

Question

A pair of charged conducting plates produces a uniform field of E = 5000 N/C, directed to the right, between the plates. The separation of the plates is 0.5 m. An electron is projected from plate A, directly toward plate B, with an initial speed of v_0 = 0.20 times 10^7 m/s. (e = 1.6 times 10^-19 C, m_electron = 9.11 times 10^-31 kg, k = 8.98 times 10^9 Nm^2/C^2) Describe the motion of the electron? What are the forces acting on electron? What is the speed of the electron as it strikes plate B? Neglect the gravity effect.

Explanation / Answer

The force due to an electric field is given by: F = qE,

Here E is the electric field and q is the charge.

however for an electron F = qE (in the opposite direction-- the electron will slow down).

From newton second law F = ma

Equating the two equation of forces

                  qE = ma

                 a = qE / m

                    = (1.602 x 10-19 C)( 5000 N/C) / (9.109 x 10-31 kg)

                    = 8.8 x 1014 m/s2

Speed of the particle from kinematic equations is

                      vf = (v02 + 2ad)
                          = ((0.2 x 107 m/s)2 + 2(8.8 x 1014 m/s2)(0.5 m))

                         = 2.973 x 107 m/s

                

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