Question

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.

Answer 1

`2.27*10^(-5) C`

`1.02*10^(-7) (N)/(C)`

Step-by-step explanation:

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

`\overrightarrow{F}=\overrightarrow{F_e}+\overrightarrow{W}`

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

`\overrightarrow{F}=\overrightarrow{F_e}+\overrightarrow{W}=0`, then

`\overrightarrow{F_e}=-\overrightarrow{W}` (1)

electric force is:

`\overrightarrow{E}=q \overrightarrow{F}` (2)

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

and weight is:

`\overrightarrow{W}=-mg` (1)

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

`q \overrightarrow{E}=mg`

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

`q (-630)=1.46*10^(-3)(9.81)`

solving for q

`q= (1.46*10^(-3)(9.81))/(-630)=2.27*10^(-5) C`

• The magnitude of weight of a proton is its mass times g:

`W=m_pg=(1.67*10^(-27))(9.81)=1.64*10^(-26)`

So electric force should be equal to W:

`F_e=1.64*10^(-26)`

`qE= 1.64*10^(-26)`

solving for E:

`E= (1.64*10^(-26))/(q)=(1.64*10^(-26))/(1.61*10^(-19))=1.02*10^(-7) (N)/(C)`