Question

What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed electric field of magnitude 700 N/C ?

Answer 1

Answer:

charge will be equal to $2.03\times 10^{-5}C$  

Explanation:

We have given mass of the particle m = 1.45 gram = 0.00145 kg

Acceleration due to gravity $g=9.8m/sec^2$

Electric field E = 700 N/C

Electric force will be equal to $F=qE$, here q is charge and E is electric field

For particle to be stationary this force must be equal to force due to gravity , that is mg force

So qE = mg

$q=\frac{mg}{E}=\frac{0.00145\times 9.8}{700}=2.03\times 10^{-5}C$

So charge will be equal to $2.03\times 10^{-5}C$