What must the charge (sign and magnitude) of a particle of mass 1.46 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 630 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 1.67×10-27 kg for the mass of a proton, 1.60×10-19 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.

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JhoanEusse JhoanEusse · Answer 1 · 2020-02-26T04:59:17+01:00

Answer:

[tex]2.27\times10^{-5} C [/tex]

[tex] 1.02\times10^{-7} \frac{N}{C} [/tex]

Explanation:

The net force (F) on the particle is the electric force (Fe) plus the weight (W) of the particle:

[tex] \overrightarrow{F}=\overrightarrow{F_e}+\overrightarrow{W}[/tex]

By Newton's first law for an object to remain stationary the net force on it should be zero, so:

[tex] \overrightarrow{F}=\overrightarrow{F_e}+\overrightarrow{W}=0[/tex], then

[tex] \overrightarrow{F_e}=-\overrightarrow{W}[/tex] (1)

electric force is:

[tex] \overrightarrow{E}=q \overrightarrow{F}[/tex] (2)

with q the charge of the particle and E the electric field

and weight is:

[tex] \overrightarrow{W}=-mg[/tex] (1)

with g the acceleration due gravity and m the mass. Using (3) and (2) on (1)

[tex]q \overrightarrow{E}=mg [/tex]

because electric field is downward directed its sign is negative so:

[tex]q (-630)=1.46\times10^{-3}(9.81) [/tex]

solving for q

[tex]q= \frac{1.46\times10^{-3}(9.81)}{-630}=2.27\times10^{-5} C [/tex]

[tex] W=m_pg=(1.67\times10^{-27})(9.81)=1.64\times10^{-26}[/tex]

So electric force should be equal to W:

[tex]F_e=1.64\times10^{-26} [/tex]

[tex]qE= 1.64\times10^{-26}[/tex]

solving for E:

[tex]E= \frac{1.64\times10^{-26}}{q}=\frac{1.64\times10^{-26}}{1.61\times10^{-19}}=1.02\times10^{-7} \frac{N}{C} [/tex]