a -21 mC charge of mass 0.47 mg is accelerated by an electric field frrom rest t

Q: a -21 mC charge of mass 0.47 mg is accelerated by an electric field frrom rest t

a) By conservation of energy (1/2)mv2 =qV (1/2)*(0.47*10-6)*(2*105)2=(21*10-3)*V V=447.6*103 Volts V=448 KV (approx) b) Fe=FB qE=mvB E=v*B =2*105*0.045 E=9000 V/m

ID: 1301567 • Letter: A

Question

a -21 mC charge of mass 0.47 mg is accelerated by an electric field frrom rest to a velocity of 2.0*10^5 m/s. it then travels into a region (shaded in the figure) that contains both an electric field and a magnetic field. the magnetic field points out of the page and has magnitude 0.045 T. The charge travels through the shaded region with a constant velocity.

a) what is the battery voltage used to accelerate the charge?
b)what is the electic field (magnitude and direction) in the shaded region?

answers are:
a) 448 kV
b) 9000 V/m up the screen

please show work/explain to get to these answers.
thanks!

Explanation / Answer

By conservation of energy

(1/2)mv2 =qV

(1/2)*(0.47*10-6)*(2*105)2=(21*10-3)*V

V=447.6*103 Volts

V=448 KV (approx)

Fe=FB

qE=mvB

E=v*B =2*105*0.045

E=9000 V/m

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a -21 mC charge of mass 0.47 mg is accelerated by an electric field frrom rest t

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