Although the changes in potential energy associated with the proton and electron

Question

Although the changes in potential energy associated with the proton and electron were similar in magnitude, the effect on their speeds differed dramatically. The change in potential energy had a far greater effect on the much lighter electron than on the proton. Question If a proton and electron both move through the same displacement in an electric field, is the change in potential energy associated with the proton equal in magnitude and opposite in sign to the change in potential energy associated with the electron? (Select all that apply.) The signs of the two changes in potential energy are opposite. They are equal in magnitude. The signs of the two changes in potential energy are the same. The magnitude of the change is larger for the proton. The magnitude of the change is smaller for the proton. PRACTICE IT Use the worked example above to help you solve this problem. A proton is released from rest at x = 2.00 cm in a constant electric field with magnitude 1.50 103 N/C pointing in the positive x-direction. (a) Assuming an initial speed of zero, find the speed of a proton at x = 0.0300 m with a potential energy of 1.20 10-17 J. (Assume the potential energy at the point of release is zero.) m/s (b) An electron is now fired in the same direction from the same position. Find the initial speed of the electron (at x = 2.00 cm) given that its speed has fallen by half when it reaches x = 0.170 m, a change in potential energy of 4.57 10-17 J. m/s EXERCISE HINTS: GETTING STARTED | I'M STUCK! The electron in part (b) travels from x = 0.170 m (where it has half the initial speed you previously calculated) to x = 0.260 m within the constant electric field. If there's a change in electric potential energy of 1.03 10-16 J as it goes from x = 0.170 m to x = 0.260 m, find the electron's speed at x = 0.260 m. (Note: Use the values from the Practice It section. Account for the fact that the electron may turn around during its travel.) vx = m/s

Explanation / Answer

1) Is the change in potential energy associated with the proton equal in magnitude and opposite in sign to the change in potential energy associated with the electron?

ANSWER :

The signs of the two changes in potential energy are opposite.

They are equal in magnitude.

Explanation :

the change in PE = V*q = E*d*q

E = electric field , d = displacement , q = charge

As the change in PE is dependent on the sign of the charge. So, the change in PE due to electron and proton is opposite in sign. but their charge being same, the magnitude will be same

2)

a) change PE = -E*d*q

= -1.5*10^3*(0.02 + 0.03)*1.6*10^-19

= -1.2*10^-17 J

change in PE = -change in KE

So, change in KE = 0.5*mv^2 = 1.2*10^-17

So, 0.5*1.67*10^-27*v^2 =1.2*10^-17

So, v = 1.2*10^5 m/s <----answer

b)

Again use the same concept, change in PE = change in KE

So, -4.57*10^-17 = -0.5*m(v^2 - (v/2)^2) <------- here initital speed = v . So, final speed = v/2

So, 4.57*10^-17 = 0.5*9.1*10^-31*(3*v^2/4)

So, v = 1.16*10^7 m/s <---------answer

c)

Again use the same concept:

change in PE = -change in KE

1.5*10^3*(2*0.17 + 0.26)*1.6*10^-19 = 0.5*9.1*10^-31(vf^2 - (3*1.16*10^7/4)^2)

So, vf = 1.98*10^7 m/s

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