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A point charge q = -0.45 nC is fixed at the origin. Where must an electron be placed in order for the electric force acting on it to be exactly opposite to its weight? (Let the y axis be vertical with the positive direction pointing up.)

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asifejaz20

Answer:

Electron must be placed at a distance of 2.69 ×10^5 m in the direction of positive y-axis.

Explanation:

Charge on electron = e = -1.6 × 10^-19 C

Mass of electron = m = 9.1 × 10^-31 kg

Charge at origin = q = -0.45 nC  

According to given condition in the question:

                       Fe = Fg

                   kq1q2/r^2 = mg  

(9 × 10^9)( -1.6 × 10^-19)( -0.45n)/r^2 = (9.1 × 10^-31)(9.8)

                    r = √7.2 ×10^10

                    r = 2.69 ×10^5 m      in the positive y-axis direction  

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