contestada

What must the charge (sign and magnitude) of a 1.45 gg particle be for it to remain stationary when placed in a downward-directed electric field of magnitude 530 n/cn/c ?

Respuesta :

The charge on the particle must be 2.68 x 10^5 C.

Electric force and Force of gravity

The given mass of the particle, m = 1.45 g = 0.00145 kg

Acceleration due to gravity = 9.8m/s²

Electric field strength, E = 530 N/C

The electric force on the particle is therefore given by;

F = qE, where,

q =charge,

F= force = mg

E is the electric field

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

Hence, qE = mg

q = mg/E.

q = (0.00145 ×9.8)/530

The charge on the particle is, therefore; 2.68 x 10^5 C.

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