Part A What must the charge (sign and magnitude) of a particle of mass 1.47 g be

Question

Part A What must the charge (sign and magnitude) of a particle of mass 1.47 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 700 N/C? Use 9.81 m/s2 for the magnitude of the acceleration due to gravity. View Available Hint(s) Submit Part B What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight? Use 1.67x10-27 kg for the mass of a proton, 1.60x10-9 C for the magnitude of the charge on an electron, and 9.81 m/s2 for the magnitude of the acceleration due to gravity View Available Hintfs N/C Submit

Explanation / Answer

a) m g = q E

q = m g / E

= (1.47 * 10-3 * 9.81) / (700)

charge of a particle = -2.06 * 10-5 C

b) E = m g / q

= (1.67 * 10-27 * 9.81) / (1.6 * 10-19)

magnitude of an electric field = 1.02 * 10-7 N/C

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