What must the charge (sign and magnitude) of a particle of mass 1.42 g be for it

Question

What must the charge (sign and magnitude) of a particle of mass 1.42 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 700 N/C? Use 9.81 m/s^2 for the magnitude of the acceleration due to gravity. What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight? Use 1.67 times 10^-27 kg for the mass of a proton, 1.60 times 10^-19 C for the magnitude of the charge on an electron, and 9.81 m/s^2 for the magnitude of the acceleration due to gravity.

Explanation / Answer

According to the question provided we need to find the value of the charge and the magntiude of electric field.

1)m g = E q ============eq(1)

q = mg / E = 0.00142 Kg * 9.8 m/s^2 / 700 = 1.988 * 10 ^ -5 C =============Part a)

2) E = 1.67 * 10^ -27 * 9.81 / 1.6 * 10^ -19 = 1.02 * 10^ -7 N/C =============Part b)

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