) A -21 mC charge of mass 0.47 mg is accelerated by an electric field from rest
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Question
) A -21 mC charge of mass 0.47 mg is accelerated by an electric field from rest to a velocity of 2.0 x 10^5 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. 1) What is the battery voltage used to accelerate the charge? 7.2) What is the electric field (magnitude and direction) in the shaded reg ion? ) A -21 mC charge of mass 0.47 mg is accelerated by an electric field from rest to a velocity of 2.0 x 10^5 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. 1) What is the battery voltage used to accelerate the charge? 7.2) What is the electric field (magnitude and direction) in the shaded reg ion?Explanation / Answer
1) conservation of energy
q V = 1/2 mv^2
21.0E-3*V = 0.5*0.47E-6*2.0E5^2
V=4.476E5
2) if going straight through
F electric = F magnetic
q E = q v B
E = v B = 2.0E5*0.045=9000
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