A uniform electric field exists in the region between two oppositely charged plane parallel plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 3.50 cmcm distant from the first, in a time interval of 2.10*10-8 ss. (a) Find the magnitude of the electric field. (b) Find the speed of the proton when it strikes the negatively charged plate.

a) E = 9.02641 * 10^-5 N/C

b) v = 3.333 * 10^6 m/s

Explanation:

Given:

- The mass of an electron m_e = 9.11 * 10^-31 kg

- The charge of an electron e = 1.602 * 10^-12 C

- Length between the plates L = 0.035 m

- Time taken t = 2.10*10^-8 s

Find:

(a) Find the magnitude of the electric field.

(b) Find the speed of the proton when it strikes the negatively charged plate.

Solution:

- The force on any object is a dependent on the acceleration of an object. Newton's Second Law of motion gives us the required expression:

F_net = m_e * a

- Assuming gravitational forces on the particle to be negligible, and only electrostatic force force is imparted on the electron as follows:

F_e = E*e

- Equating the above two equations we have:

E = m_e*a / e

- The acceleration of an electron in the Electric field E can be determined from the the second kinematic equation of motions. Where initial distance x_o = 0 and initial velocity v_x, o = 0. Where x is the direction along the distance between the plates L. We have:

x (f) = L = 0.5*a*t^2

a = 2*L / t^2

- Substituting the expression for acceleration a into the previously derived expression for Force, we get:

E = 2*m_e * L / e*t^2

- Plug in the values and solve for E:

E = 2 * (9.11*10^-31) * (0.035) / (1.602*10^-12) * (2.1*10^-8)

E = 6.377 * 10^-32 / 7.06482*10^-28

E = 9.02641 * 10^-5 N/C

- For velocity we can use our relationship for acceleration:

v = a*t = 2*L / t

v = 2 * (0.035) / (2.1*10^-8)

v = 3.333 * 10^6 m/s