Answer:
a) E = 9.02641 * 10^-5 N/C
b) v = 3.333 *10^6 m/s
Step-by-step 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