A proton accelerates from rest in a uniform electric field o

A proton accelerates from rest in a uniform electric field of 640 N/C. At some time later, its speed is 120 times 10^6 m/s. The mass of a proton is 1.673 times 10^-17 kg and the charge of a proton is 1.6 times 10^-19 C. What is the magnitude of the acceleration of the proton? How long does it take the proton to reach this speed? How far has it moved in this time interval? What is its kinetic energy at the later time? An electric field of magnitude 3500 N/C is applied along the y axis. Calculate the electric flux through a rectangular plane 0.35 m wide and 0.7 m long if the plane is parallel to the yz plane. Calculate the electric flux through the same rectangle, if it is parallel to the xy plane. Calculate the electric flux through the same rectangle, but now the rectangle contains the y axis and its normal makes an angle of 40 with the x axis.. All the shaded areas represent conductors, and the un-shaded areas represent air. Assume that the outer sphere (Area of a sphere is 4nr2) has a total charge of -5.0 \\iC. The inner sphere has a charge of -2.0 muC. What is the charge on the inner surface of the bigger sphere? What is the charge on the outer surface of the bigger sphere? What is the electric field at A (radius = 0.02m)? What is the electric field at B (radius = =0.55m)?

Solution

6)

E = 640 N/C

v = 1.2 *10^6 m/s

mp = 1.673 *10^-27 Kg

q = 1.6 *10^-19 C

a)

let the magnitude of acceleration is a

m * a = E * q

1.673 *10^-27 * a = 640 * 1.6 *10^-19

a = 6.13 *10^10 m/s^2

the acceleration of the proton is 6.13 *10^10 m/s^2

b)

let the time taken is t

1.2 *10^6 = 6.13 *10^10 * t

t = 1.96 *10^-5 s

the time taken is 1.96 *10^-5 s

c)

distance travelled = 0.5 * a * t^2

distance travelled = 0.5 *6.13 *10^10* (1.96 *10^-5)^2

distance travelled = 11.8 m

d)

kinetic energy at later time = 0.5 * m * v^2

kinetic energy at later time = 0.5 * (1.2 *10^6)^2 *1.673 *10^-27

kinetic energy at later time = 1.205 *10^-15 J