What must the charge (sign and magnitude) of a 1.70 g particle be for it to remain balanced against gravity when placed in a downward-directed electric field of magnitude 700 n/c ?

Charge,

Explanation:

Mass of the particle, m = 1.7 g = 0.0017 kg

Electric field, E = 700 N/C

We need to find the charge particle to  remain balanced against gravity when placed in a downward-directed electric field. So,

q E = m g

q = 0.0000238 C

So, the magnitude of charged particle is . As the direction of electric field is in downward direction, so the charge is negative. Hence, this is the required solution.

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explanation:

graphing motion: mastery test

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velocity imetersisecond

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time (seconds)

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