B. 8
C. 4
D. 12
C. 4 beats per second is correct
(b) Find the magnitude and direction of the electric field that the particle experiences. _______N/C the direction of motion (along, against, perpendicular)
The magnitude of the electric field is 4000 N/C while the direction of the electric field is along the direction of the motion of the charge.
Given to us
Charge of the particle, q = +1.5 µC = 1.5 x 10⁻⁶
Distance between points A and B, d = 0.15 m
The particle's electric potential energy at A and B
= = +9.0x10⁻⁴ J
The electric charge in electric potential energy is equal to the work done by an electric field. therefore,
W = 9.0x10⁻⁴ J
We know that work done by the system can be given as,
Substitute the values,
F = 1.5 x 10⁻³ N
Thus, the magnitude of the force is 1.5 x 10⁻³ N while its direction is along with the motion of the charge.
We know the relationship between electrostatic and electric force, therefore,
Substitute the values,
E = 4000 N/C
Hence, the magnitude of the electric field is 4000 N/C while the direction of the electric field is along the direction of the motion of the charge.
Learn more about the Electric field:
Answer:
Part a)
Direction of force is along the motion of charge
Part b)
direction of electric field is along the direction of motion
Explanation:
Part a)
As we know that the change in electric potential energy is equal to the work done by electric field
now from the equation of work done we know that
Direction of force is along the motion of charge
Part b)
As we know the relation between electrostatic force and electric field given as
direction of electric field is along the direction of motion
B) The alpha particle will reverse its direction.
C) The alpha particle will be deflected in a curve path.
Eliminate
D) The alpha particle will continue to travel in a straight line.
Answer:
C. The alpha particle will be deflected in a curve path.
Reflection is the answer