Physics: Example Problem and Solution
Not what you're looking for?
A charged proton particle is moving rightward between two parallel charged plates separated by distance d= 6.60mm. The plate potentials are V1= -67.0 v and V2= -48.0v. The particle is slowing from an initial speed of 97.0 km/s at the left plate. What is its speed just as it reaches plate 2?
A thin spherical, conducting shell of radius R is mounted on an isolating support and charged to a potential of -799v. An electron is then fired directly toward the center of the shell, from point P at distance r from the center of the shell, ( r is greater than R). What initial speed vo is needed for the electron to just reach the shell before reversing direction?
What is the electron potential energy in joules of of two electrons separated by 2.33nm? And what would it be if the separation was doubled?
In a shape of a square is a charge of four particles on each corner of the square.
In the order of positive, negative, positive, negative. How much work in joules is required to set up the four charge configuration if q= 1.62pC, a= 77.1cm, and the particles are initially infinitely far apart and at rest?
Purchase this Solution
Solution Summary
This solution provides steps necessary to determine the answers to the physics questions attached.
Solution Preview
Please see the attached.
A charged proton particle is moving rightward between two parallel charged plates separated by distance d= 6.60mm. The plate potentials are V1= -67.0 v and V2= -48.0v. The particle is slowing from an initial speed of 97.0 km/s at the left plate. What is its speed just as it reaches plate 2?
The potential difference between two points is the difference in the electrostatic potential energy per unit charge. Thus the change in the potential energy of the proton (charge +e) when it moves from plate 1 to plate 2 is given by
∆U_E= +e(V_2-V_1 )
As there is no other non conservative force acting on the proton, the total energy remains conserved and thus according to law of conservation of energy we can write Loss in kinetic energy = gain in electrostatic potential energy
Or 1/2 mv_1^2-1/2 mv_2^2= +e(V_2-V_1 )
Or v_1^2-v_2^2= +2e(V_2-V_1 )/m
Here m is the mass of the proton.
Thus speed of the proton when it just reaches plate 2 is given by
v_2^2= v_1^2-2e(V_2-V_1 )/m
Or v_2= √(v_1^2-2e(V_2-V_1 )/m)
Now the initial speed v_1=97.0 km/s = 9.70*〖10〗^4 m/s
Charge on proton e=+1.6*〖10〗^(-19) C
Mass of proton m=1.67*〖10〗^(-27) kg
Thus ...
Purchase this Solution
Free BrainMass Quizzes
The Moon
Test your knowledge of moon phases and movement.
Variables in Science Experiments
How well do you understand variables? Test your knowledge of independent (manipulated), dependent (responding), and controlled variables with this 10 question quiz.
Classical Mechanics
This quiz is designed to test and improve your knowledge on Classical Mechanics.
Intro to the Physics Waves
Some short-answer questions involving the basic vocabulary of string, sound, and water waves.
Basic Physics
This quiz will test your knowledge about basic Physics.