7) A -21 mC charge of mass 0.47 mg is accelerated by an electric field from rest

Question

7) A -21 mC charge of mass 0.47 mg is accelerated by an electric field from rest to a velocity of 2.0 x
105 m/s. It then travels into a region (shaded in figure) that contains both an electric field and a
magnetic field. The magnetic field points out of the page and
has a magnitude of 0.045 T. The charge travels through the
shaded region with a constant velocity.
7.1) What is the battery voltage used to accelerate the
charge?
7.2) What is the electric field (magnitude and direction) in the
shaded region?

I know at some point I need to use conservation of energy but I'm lost as to how to start the problem.

Thanks

Explanation / Answer

7.1 conservation of energy

eV = 1/2 mv^2

2.1E-3*V = 0.5*0.47E-6*2.0E5^2

V=4.47E6 V

7.2 to go straight, Fmagnetic = F electric

q v B = q E

E = v B = 2.0E5*0.045= 9000 N/C

magnetic force will point up so E will point up since q is negative to make electric force down

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