What must the charge (sign and magnitude) of a particle of mass 1.44 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 660 n/c ? use 9.81 m/s^2 for the magnitude of the acceleration due to gravity.

Charge,

Explanation:

It is given that,

Mass of the charged particle, m = 1.44 g = 0.00144 kg

Electric field, E = 660 N/C

We need to find the charge of that particle to remain stationary when placed in a downward-directed in the given electric field such that its weight is balanced by the electrostatic force i.e.

q = 0.0000214 C

Since, the electric field is acting in downward direction, so the electric force will act in opposite direction such that they are in balanced position. Hence, the charge must be negative.

i.e.

real, smaller and upside-down.   so the answer should be b.

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west