problems. Your score will be based on the maximum grade a otherwise. Number of c

Question

problems. Your score will be based on the maximum grade a otherwise. Number of choices in mutiple choice question is ether one or two depending upon the number of choices submissions remaining for each question part only changes f you submt or change the arswer got in this test. You get tw for each question, uniess stated Your last submission is used for your score capacitor at an angle of 53 through a small hole in the bottom plate as shown in the figure below. If the plates are Aproton with an initial velocity of 3.2 x 106 m/s enters a by a distance d 2.3 cm, what is the magnitude of the minimum electric field to prevent the proton from hitting the top plate? N/C Compare your answer with the results obtained if the proton was replaced with an electron The magnitude of the electric field needed for a proton is smaller than what is needed for an electron The magnitude of the electric field needed for a proton is the same as what is needed for an electron The magnitude of the electric field needed for a proton is larger than what is needed for an electron. 2 3

Explanation / Answer

velocity perpendicular to the plates,

v0y =(3.2 x 10^6 m/s) sin53 = 2.56 x 10^6 m/s

Suppose after travelling distance d in this direction, its velocity becomes zero.

so applying vF^2 - vi^2 = 2 a d

0^2 - (2.56 x 10^6)^2 = 2(a)(0.023)

a = 1.42 x 10^14 m/s^2

and a = F / m = q E / m

E = m a / q

E = (1.67 x 10^-27) (1.42 x 10^14) / (1.6 x 10^-19)

E = 1.48 x 10^6 N/C .......Ans

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E = m a / q

mass of electron is smaller than proton.

hence

magnitude of field for proton is larger than what is needed for electron.

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