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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?

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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 switching in the values in equation above we get
v_2= √((9.70*〖10〗^4 )^2-(2*1.6*〖10〗^(-19) [-48.0-(-67.0)])/(1.67*〖10〗^(-27) ))
Or v_2= √(9.41*〖10〗^9-3.64*〖10〗^9 )=7.60*〖10〗^4 m/s
Thus the speed of the proton just before it reaches the second plate is 76.0 km/s.

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?
The potential due to a charged conducting sphere of radius R and charge Q at a distance r (≥R) is given by
V_r=Q/(4πϵ_0 r)
As the potential at the surface of the sphere is
V_R=Q/(4πϵ_0 R)= -799 V
And V_r=Q/(4πϵ_0 r)=Q/(4πϵ_0 R) (R/r)= -799(R/r)
As the electron is just reached the shell before reversing the direction, its velocity and hence the kinetic energy at the surface is zero, and thus we can say that its kinetic energy is lost to gain in potential energy.
Thus Loss in kinetic energy = gain in electrostatic potential energy
Or 1/2 mv_0^2-0=e*∆V
Or 1/2 mv_0^2=e*(V_R-V_r )
Or v_0=√((2e*(V_R-V_r ))/m)
Substituting the values we get
v_0=√((2*(-1.6*〖10〗^(-19) )*[-799-(-799)(R/r)])/(9.11*〖10〗^(-31) ))=1.68*〖10〗^7 √(1-R/r) m/s
If R << r, R/r will be negligibly small and thus the speed of the electron to reach the surface of the sphere will be 1.68*107 m/s.

What is the electron potential energy in joules of two electrons separated by 2.33nm? and what would it be if the separation was doubled?

The electrostatic potential energy of a system of two point charges q1 and q2 at a distance r is given by
U_E=(q_1 q_2)/(4πϵ_0 r)=9*〖10〗^9*(q_1 q_2)/r
Hence the energy of the electrons is given by
U_1=9*〖10〗^9*((-1.6*〖10〗^(-19) )*(-1.6*〖10〗^(-19) ))/(2.33*〖10〗^(-9) )=9.89*〖10〗^(-20) J
Now if the separation between the electrons will be doubled, the energy of the system will be halved and hence
U_2=1/2*9.89*〖10〗^(-20)=4.94*〖10〗^(-20) J
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?
The electrostatic energy of a system of point charges is given by
U=1/2 ∑_(i=1)^n▒〖 (q_i V_i ) 〗
Here qi is one charge and Vi is the potential at the position of qi due to all other charges. Thus for the system of four charges on the vertices of a square of side a is given by
U=1/2 (q_1 V_1+q_2 V_2+q_3 V_3+q_4 V_4 )
Or U=1/2 [█(+q ((-q)/(4πϵ_0 a)+(+q)/(4πϵ_0 √2 a)+(-q)/(4πϵ_0 a))+(-q) (q/(4πϵ_0 a)+(-q)/(4πϵ_0 √2 a)+q/(4πϵ_0 a))@+q ((-q)/(4πϵ_0 a)+(+q)/(4πϵ_0 √2 a)+(-q)/(4πϵ_0 a))+(-q) (q/(4πϵ_0 a)+(-q)/(4πϵ_0 √2 a)+q/(4πϵ_0 a)) )]
Or U=1/2 [4 q ((-q)/(4πϵ_0 a)+(+q)/(4πϵ_0 √2 a)+(-q)/(4πϵ_0 a))]
Or U=(2q^2)/(4πϵ_0 a) (-2+1/√2)
Substituting values we get
U=9*〖10〗^9*(2*(1.62*〖10〗^(-12) )^2)/(77.1*〖10〗^(-2) ) (-2+1/√2)=-7.92*〖10〗^(-14) J
The energy of the charges when they are at infinite distance is zero hence the work done to setup the configuration is - 7.92*10-14 J.
As the work done is negative, this means we have to do negative work or the work done by the system of charges. The energy is released by the system.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!