標題:
PHY The electric field
發問:
1. A proton of mass 1.67*10^-27 enters the region between 2 parallel plates a distance 20cm apart. There is a uniform electric field of 2*10^4V/m between the plates. If the initial speed of the proton is 7*10^6m/s what is the final speed?2. Two protons (mass=1.67*10^-27kg, charge=1.6*10^-19C) in a nucleus are... 顯示更多 1. A proton of mass 1.67*10^-27 enters the region between 2 parallel plates a distance 20cm apart. There is a uniform electric field of 2*10^4V/m between the plates. If the initial speed of the proton is 7*10^6m/s what is the final speed? 2. Two protons (mass=1.67*10^-27kg, charge=1.6*10^-19C) in a nucleus are 10^-15m apart. a) what is their electrial potential energy? b) If they were free to move and start from rest, find their speeds when they are 5*10^-15m apart. 3. External work equal to 5*10^-7J is needed to move a -6nC charge at constant speed to a point at which the potential is -20V. What is the potential at the initial point? 4. Referring to the diagram a) what is the magnitude of the electric field at Q2? b) what is the potential at Q2? c) what is the potential energy of Q2? 圖片參考：http://imgcld.yimg.com/8/n/HA07982681/o/701112280033113873416090.jpg 5. Referring to the diagram a) what is the mannitude of the electric field at Q2? b) what is the potential at Q2? c) what is the potential energy of Q2? 圖片參考：http://imgcld.yimg.com/8/n/HA07982681/o/701112280033113873416091.jpg
1. acceleration = 1.6x10^-19 x 2x10^4/1.67x10^-27 m/s2 = 1.92 x 10^12 m/s2 final velocity = square-root[(7x10^6)^2 + 2 x 1.92x10^12 x 0.2] m/s = 7.05x10^6 m/s 2. (a) Electrical potential energy = (9x10^9) x (1.6x10^-19)^2/10^-15 J = 2.304 x 10^-13 J (b) Potential energy when the protons are 5x10^-15 m apart = (9x10^9) x (1.6x10^-19)^2/5x10^-15 J = 4.608x10^-14 J loss of potential energy = (2.304x10^-13 - 4.608x10^-14) J = 1.84x10^-13 J By conservation of energy, speed of proton = square-root[2 x 1.84x10^-13/1.67x10^-27] m/s = 1.48x10^7 m/s 3. Potential difference between the two points = 5x10^-7/6x10^-9 v = 83.3 v Hence, potential of the initial point = (-20 + 83.3) v = 63.3 v 4. (a) Electirc field = (9x10^9).[30x10^-6/0.1^2 - 8x10^-6/0.05^2] N/C = 2.69x10^7 N/C (b) Potential = (9x10^9).[30x10^-6/0.1 + 8x10^-6/0.05] v = 4.14x10^6 v (c) Potential energy of Q2 = 4.14x10^6 x (-5x10^-6) J = -20.7 J 5. (a) Field due to Q1 = (9x10^9) x 8x10^-6/0.03^2 N/C = 8x10^7 N/C Field due to Q3 = (9x10^9) x 8x10^-6/0.04^2 N/C = 4.5x10^7 N/C Hence, resultant field = square-root[(8x10^7)^2 + (4.5x10^7)^2] N/C = 9.18x10^7 N/C (b) Potential = (9x10^9).[8x10^-6/0.03 + 8x10^-6/0.04] v = 4.2x10^6 v (c) Potential energy of Q2 = 4.2x10^6 x (-6x10^-6) J = -25.2 J
其他解答:4E350C6F8B48ECA2
PHY The electric field
發問:
1. A proton of mass 1.67*10^-27 enters the region between 2 parallel plates a distance 20cm apart. There is a uniform electric field of 2*10^4V/m between the plates. If the initial speed of the proton is 7*10^6m/s what is the final speed?2. Two protons (mass=1.67*10^-27kg, charge=1.6*10^-19C) in a nucleus are... 顯示更多 1. A proton of mass 1.67*10^-27 enters the region between 2 parallel plates a distance 20cm apart. There is a uniform electric field of 2*10^4V/m between the plates. If the initial speed of the proton is 7*10^6m/s what is the final speed? 2. Two protons (mass=1.67*10^-27kg, charge=1.6*10^-19C) in a nucleus are 10^-15m apart. a) what is their electrial potential energy? b) If they were free to move and start from rest, find their speeds when they are 5*10^-15m apart. 3. External work equal to 5*10^-7J is needed to move a -6nC charge at constant speed to a point at which the potential is -20V. What is the potential at the initial point? 4. Referring to the diagram a) what is the magnitude of the electric field at Q2? b) what is the potential at Q2? c) what is the potential energy of Q2? 圖片參考：http://imgcld.yimg.com/8/n/HA07982681/o/701112280033113873416090.jpg 5. Referring to the diagram a) what is the mannitude of the electric field at Q2? b) what is the potential at Q2? c) what is the potential energy of Q2? 圖片參考：http://imgcld.yimg.com/8/n/HA07982681/o/701112280033113873416091.jpg
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最佳解答:1. acceleration = 1.6x10^-19 x 2x10^4/1.67x10^-27 m/s2 = 1.92 x 10^12 m/s2 final velocity = square-root[(7x10^6)^2 + 2 x 1.92x10^12 x 0.2] m/s = 7.05x10^6 m/s 2. (a) Electrical potential energy = (9x10^9) x (1.6x10^-19)^2/10^-15 J = 2.304 x 10^-13 J (b) Potential energy when the protons are 5x10^-15 m apart = (9x10^9) x (1.6x10^-19)^2/5x10^-15 J = 4.608x10^-14 J loss of potential energy = (2.304x10^-13 - 4.608x10^-14) J = 1.84x10^-13 J By conservation of energy, speed of proton = square-root[2 x 1.84x10^-13/1.67x10^-27] m/s = 1.48x10^7 m/s 3. Potential difference between the two points = 5x10^-7/6x10^-9 v = 83.3 v Hence, potential of the initial point = (-20 + 83.3) v = 63.3 v 4. (a) Electirc field = (9x10^9).[30x10^-6/0.1^2 - 8x10^-6/0.05^2] N/C = 2.69x10^7 N/C (b) Potential = (9x10^9).[30x10^-6/0.1 + 8x10^-6/0.05] v = 4.14x10^6 v (c) Potential energy of Q2 = 4.14x10^6 x (-5x10^-6) J = -20.7 J 5. (a) Field due to Q1 = (9x10^9) x 8x10^-6/0.03^2 N/C = 8x10^7 N/C Field due to Q3 = (9x10^9) x 8x10^-6/0.04^2 N/C = 4.5x10^7 N/C Hence, resultant field = square-root[(8x10^7)^2 + (4.5x10^7)^2] N/C = 9.18x10^7 N/C (b) Potential = (9x10^9).[8x10^-6/0.03 + 8x10^-6/0.04] v = 4.2x10^6 v (c) Potential energy of Q2 = 4.2x10^6 x (-6x10^-6) J = -25.2 J
其他解答:4E350C6F8B48ECA2
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