what must the charge (sign and magnitude) of a 1.48-g particle be for it to remain stationary when placed in a downward-directed electric field of magnitude 660 n/c ? you may want to review (page) .



Answer :

Mass of the particle m=1.48 g=1.48×10−3 kgm=1.48 g=1.48×10−3 kg

Electric field E=670 N/CE=670 N/C directed downwards.

Charge Particle in an Electric Field

The electric force must act as opposite to the weight of the particle

qE= mgxqE = mg

q = mgEq= mgE

q= 1.48×10−3  kg × 9.8 m/s 2670 N/C

q=2.16×10−5 Cq=2.16×10−5 C

As the force is upwards and an electric field is downwards it means the charge on the particle is negative.

q=−2.16 × 10−5 Cq= −2.16 × 10−5 C

In the second case for a proton

qE= mgqE= mg

E= mgqE= mgq

E= 1.67×10−27 kg × 9.8 m/s21.6×10−19 C

E= 1.023×10−7 N/C

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