The drawing shows a uniform electric field that points in the negative y directi

Question

The drawing shows a uniform electric field that points in the negative y direction: the magnitude of the field is 5900 N/C. A 15 mg particle with 100 nC of charge moves from B to C under the influence of this field. A different particle of same mass and charge moves from A to C within the same field. Potential difference between points A and B is _____ Potential difference between points B and C is _____ Potential difference between points A and C is _____ The speed gained by particle which moved from B to C _____ The speed gained by the particle which moved from A to C _____

Explanation / Answer

2)Given,

E = 5900 N/C ; m = 15 mg ; q = 100 nc ;

a)The potential difference between A and B is zero. Since they lie at the same equipotential line.

Vab = 0

b)The potentil difference between B and C will be:

Vbc = E d

Vbc = 5900 x 0.08 = -472 Volts

Hence, Vbc = -472 Volts

c)Vca = 472 V

d)from energy conservation

KE = PE

1/2 m v^2 = q V

v = sqrt (2 q V/m)

v = sqrt (2 x 100 x 10^-6 x 472/15 x 10^-6) = 79.33 m/s

Hence, v = 79.33 m/s

e)velocity will be same as (d) since V is same.

v = 79.33 m/s

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