Question

A proton is traveling horizontally to the right at 4.00×106 m/s. (a) Find the magnitude and direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 1.67 cm. (b) How 7 much time does it take the proton to stop after entering the field?A −1.00-nC point charge is at the origin, and a +4.00-nC point charge is on the y-axis at y = 2.00 m. (a) Find the electric field (magnitude and direction) at each of the following points on the y-axis: (i) y = 3.00 m; (ii) y = 1.00 m; (iii) y = −2.00 m. (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).

Answer 1

Mass of proton

Charge of proton

Initial velocity of proton

Final velocity of proton

Displacement of proton

Acceleration of proton

(a)

Using the Work-Energy theorem, work done by all forces is change in kinetic energy.

Magnitude of electric field

Direction of electric field is to the left.

(b)

Using the equation

From above equation

____________________________________________________________

A point charge is located at point . Electric field at point due to the charge is

located at

located at

a)

i)

Electric field at due to the two charges is

Magnitude of electric field is

Direction of electric field is towards the positive Y-axis.

ii)

Electric field at   due to the two charges is

Magnitude of electric field is

Direction of electric field is towards the negative Y-axis.

iii)

Electric field at due to the two charges is

Magnitude of electric field is

Direction of electric field is none.

b)

Net electric force on an electron placed at is

Magnitude of net electric force is and direction of net eletric force is towards -ve Y-axis.

Net electric force on an electron placed at is

Magnitude of net electric force is and direction of net eletric force is towards +ve Y-axis.

Net electric force on an electron placed at is

Magnitude of net electric force is and direction of net eletric force not defined.