Three charged metal screens that are in an evacuated pipe (no air inside the pip

Question

Three charged metal screens that are in an evacuated pipe (no air inside the pipe). The electric potential of the screens is as follows: V1= 100·V, V2= 300·V, and V3= 0·V. A small particle with a charge of q = 0.002·C is moving horizontally through the pipe from left to right. When it reaches the first screen, the particle has a kinetic energy of KE = 1·J. particle in pipe (a) What is the potential energy of the particle when it reaches the first screen? J (b) What is the mechanical energy of the particle when it reaches the first screen? J (c) What is the potential energy of the particle when it reaches the second screen? J (d) What is the kinetic energy of the particle when it reaches the second screen? J (e) What happens to the speed of the particle as it moves from the first to the second screen? Choose one answer only. increases remains constant decreases (f) What is the potential energy of the particle when it reaches the third screen? J (g) What is the kinetic energy of the particle when it reaches the third screen? J (h) What happens to the speed of the particle as it moves from the second to the third screen? Choose one answer only. increases remains constant decreases

Explanation / Answer

Solution: Electric Potential at screen 1, V1 = 100 V

Electric Potential at screen 2, V2 = 300 V

Electric Potential at screen 3, V3 = 0 V

Charge on the particle q = 0.002 C

Kinetic energy of the charge q at the screen 1 = 1 J

Part (a) The electric potential at the screen 1 is 100 V (or 100 J/C). Thus the electric potential energy U1 at the screen 1 is given by,

U1 = q*V1

U1 = (0.002C)*(100V)

U1 = (0.002C)*(100J/C)

U1 = 0.2 J

Thus the answer is 0.2 J

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Part (b) When the particle is at screen 1, it has kinetic energy K1 = 1 J and as calculated above it has electric potential energy U1 = 0.2 J.

The total mechanical energy of the particle is,

T1 = U1+ K1

T1 = 0.2 J + 1.0 J

T1 = 1.2 J

Thus the answer is 1.2 J

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Part (c) When the particle reaches the screen 2, it has the electric potential as that of the screen 2; that is V2 = 300 V.

Thus the electric potential energy U2 at the screen 2 is given by,

U2 = q*V2

U2 = (0.002C)*(300V)

U2 = (0.002C)*(300J/C)

U2 = 0.6 J

Thus the answer is 0.6 J

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Part (d) When the particle reaches second screen, it has electric potential energy of 0.6 J.

But we know that the total mechanical energy of the particle must remain same during the entire motion. That is T2 = T1 = 1.2 J

T2 = K2 +U2

1.2 J = K2 + 0.6 J

K2 = 0.6 J

Thus the answer is 0.6 J

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Part (e) Kinetic energy of the particle is K = (1/2)*m*v2 where is m = mass and v = speed of the particle.

When kinetic energy of the particle increases, its speed increases and vice versa.

As the particle moves from first screen to second screen, its kinetic energy decreases from 1.0 J to 0.6 J,

thus the speed of the particle decreases.

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Part (f) The electric potential at the screen 3 is 0 V (or 0 J/C). Thus the electric potential energy U3 at the screen 3 is given by,

U3 = q*V3

U3 = (0.002C)*(0V)

U3 = (0.002C)*(0J/C)

U3 = 0.0 J

Thus the answer is 0.0 J

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Part (g) Total mechanical energy remains the same, That is T3 =T2 = T1 = 1.2 J

T3 = K3 +U3

1.2 J = K2 + 0.0 J

K2 = 1.2 J

Thus the answer is 1.2 J

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Part (h) As the particle moves from second screen to third screen, its kinetic energy increases from 0.6 J to 1.2 J,

thus the speed of the particle increases.

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