f +2e and a mass of6.6×10-27 kg, what is the magnitude ofthe electric force exac
ID: 1731559 • Letter: F
Question
f +2e and a mass of6.6×10-27 kg, what is the magnitude ofthe electric force exactly balances the weight of the helium nucleus so that it nelium nucleus is located between the plates of a parallel-plate capacitor as shown. The nucleus has a chargeo field such that the electric remains stationary? +2e 772 A) 4.0 x 10 N/C B) 6.6 x 10-26 N/C C) 2.0× 10-7 NC D) 5.0x 10-3 N/C E) 1.4×10-8 NC 12. The of a -3.0 uC charge placed at that point? electric potential at a certain point is space is 12 V. What is the electric potential energy B) -4 uJ C) +36 pu D)-36 J E) zero ?J 13. A completely ionized beryllium atom (net charge +4e)is accelerated through a potential difference of 6.0 V. What is the increase in kinetic energy of the atom? A) zero ev B) 0.67 eV C) 4.0 eV D) 6.0 eV E) 24 eV point charges are arranged along the x axis as shown in the figure. At which of the 14. Two following values of x is the electric potential equal to zero? Note: At infinity, the electric potential is zero. +2.0 JIC -5.0 C 1.0m A) +0.05 m B) +0.29 m C) +0.40 m D) +0.54 m E) +0.71 mExplanation / Answer
11)
let the electric field be E
use:
electric force = weight
q*E = m*g
(2e)*E = (6.6*10^-27 Kg)*(9.8 m/s^2)
(2*1.602*10^-19)*E = (6.6*10^-27 Kg)*(9.8 m/s^2)
E = (6.6*10^-27)*(9.8)/(2*1.602*10^-19)
= 2.0*10^-7 N/C
Answer: C
12)
EPE = q*V
= (-3.0 uC)*(12 V)
= -36 uJ
Answer: D
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