A small particle has charge -6.30 C and mass 1.50×104 kg . It moves from point A

Question

A small particle has charge -6.30 C and mass 1.50×104 kg . It moves from point A, where the electric potential is VA = 300 V , to point B, where the electric potential VB = 540 V . The electric force is the only force acting on the particle. The particle has a speed of 3.20 m/s at point A.

What is its speed at point B?

Express your answer in meters per second to three significant figures.

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Part B

Is it moving faster or slower at B than at A?

Is it moving faster or slower at  than at ?

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Explanation / Answer

charge =-6.3x10-6C

mass = 1.5x10-4kg

initial velocity =3.2m/s

final velocity= ?

Change in kinetic energy = change in potential energy

(1/2)mv2 - (1/2)mu2 = q (Va - Vb)

multiply by 2 and divide by m on both sides

v2 - u2 = 2q (Va - Vb) /m

v2 = u2 + 2q (Vb - Va) /m

v = sq rt [ u2 + 2q (Va - Vb) /m ]

v = sq rt [ 3.22+ 2x(-6.3x10-6) (300 -540) / 1.50×104 ]

v = sq rt [ 10.24 + 20.16 ]

v = sq rt [ 30.4 ]

v = 5.5 m/s

A) Its speed at point B is 5.5m/s

B)It is moving faster at B than at A

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