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12. An object with a 5.0 HC charge is accelerating at 0.0050 m/s2 due to an electric field. The( potnt) object has a mass of 2.0 mg. What is the magnitude of the electric field? O2.0 N/C O-0.0020 N/C O0.0020 NO O-2ONC 13. Point P and point charge are separated by a distance R. The electric field at point P has (1 point) magnitude E. How could the magnitude of the electric field at point P be doubled to 2E? Odouble the distance to 2R Oreduce the distance to R/2 Oreduce the distance to R/4 Oreduce the point charge to 9/2 Odouble the point charge to 20

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Answer #1

12) acceleration =Net force / mass

a = qE/m

where q = charge , E = electric field

m = mass of the object

0.005 = 5*10^-6*E / 2*10^-6

E = 0.002 N/C

13) Electric field = kQ/r^2

so electric field at P can by doubled by doubling the charge Q

so option double the point charge 2Q is the correct answer

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