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An alpha particle(the nucleus of a helium atom) has a mass of 6.64*10^-27kg and a charge of +2e. What are the magnitude and direction of the electric field that will balance the gravitational force on it? Mp= 1.67*10^-27kg

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Weight = electric force 
mg = qE 

6.64x10^-27 x 9.81 = (2 x 1.60x10^-19) E 
qE =mg, 
E = mg/q = 6.64•10^-27•9.8/2•1.6•10^-19 =2.03•10^-7 V/m
anniwuanyanwu1

Answer:

Magnitude of electric field E= 2.03×10^-7NC^-1

The charge is positive,the electric force is in the same direction as the electric field according to F= qE

Therefore,the direction of the electric field is upward.

Explanation:

Fg=Fe

Where Fe is the force of the electric field

Fg is force of the gravitational field.

Mg= we

Mg= 2eE

E=mg/2e

E= (6.64×10^-27)(9.8)/2(1.6×10^-19)

E=( 6.50×10^-26)/(3.2×10^-19)

E= 2.03×10^-7NC^-1

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