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If only electric forces are important Conservation of Mechanical Energy tells us

ID: 1399191 • Letter: I

Question

If only electric forces are important Conservation of Mechanical Energy tells us . In practice the electric force is frequently much larger than the gravitational force, and we can eliminate frictional or drag forces by using a vacuum system. The the only potential energy is the electrostatic potential energy, U W. An electron has a speed of 5.2 x 10^5 m/s at point A. The voltage at point A is 3500 V and at point B is 4775 V. (a) What is the speed of the electron at point B? Assume forces other than the electric force are negligible. (b) Repeat the question for a proton instead of an electron. Notice the big difference between accelerating an electron and a proton.

Explanation / Answer

Given that,

velocity at A = v1= 5.2 x 105 m/s ; Voltage at A = V1 = 3500 Volts

Now voltage at point B V2 = 4775 Volts we need to find the velocity of electron at B let it be v2.

We know that,

qV = 1/2 mv2

qV1 = 0.5 mv12 (for region A)

qV2 = 0.5 m v22 (for region B)

dividing the above equations we get:

V1/V2 = v12/v22 => v2 = v1 x sqrt(V2/V1)

v2 = 5.2 x 105 x sqrt (4775/3500) = 2.92  x 105 m/s

Hence, v2 =  2.92  x 105 m/s.

(b)for proton we know that, magnitude of charge is same but there is a diffrence of mass. So at V1 = 3500 V, velocity of proton will be:

qV1 = 0.5 mv12 => v1 = sqrt(2 q V1/ m)

v1 = sqrt (2 x 1.6 x 10-19 C x 3500 / 1.67 x 10-27 )

v1 = 81.89x 104 = 8.18 x 105 m/s

v2 = v1 x sqrt(V2/V1) = 8.18 x 105 x sqrt(4775/3500) = 9.55 x 105 m/s

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