# Physics: Example Problem and Solution

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

âˆ†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= âˆš(v_1^2-2e(V_2-V_1 )/m)

Now the initial speed v_1=97.0 km/s = 9.70*ã€–10ã€—^4 m/s

Charge on proton e=+1.6*ã€–10ã€—^(-19) C

Mass of proton m=1.67*ã€–10ã€—^(-27) kg

Thus ...

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