The main concepts used to solve this problem are the electric field, force, and kinematic equation for speed.
Initially, calculate the acceleration of the object by using the equation of motion. Later calculate the electric field by using the relationship between the force and charge and finally calculate the speed of proton.
The force acting on the object due to the Newton’s second law is,
Here, m is the mass and a is the acceleration of the object.
The expression for the force acting between two oppositely charged plane parallel plates is,
Here, q is the charge on the plates and E is the electric field between the plates.
Equate the above two expression of the forces and simplify the expression for the electric field.
The kinematic equation for the speed is,
Here, u is the initial speed, v is the final speed, x is the distance, and a is the acceleration.
The kinematic equation for the distance is,
Here, x is the distance, u is the initial speed, t is the time, and a is the acceleration.
(A)
The kinematic equation for the distance is,
Substitute 0 for u and rearrange the equation for the a.
Substitute 1.50 cm for x and for t.
The expression for the electric field is,
Substitute for m, for a, and for q.
(B)
The kinematic equation for the speed is,
Here, u is the initial speed, v is the final speed, x is the distance, and a is the acceleration.
Initially, the proton is at rest on the surface.
Substitute 0 m/s for u in the equation and solve for .
The expression for the speed of the proton is,
Substitute for a and 1.50 cm for x.
Ans: Part A
The magnitude of the electric field is 157.5 N/C.
A uniform electric field exists in the region between two oppositely charged parallel plates 1.50 apart....
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