a = F/m = keQ1Q2/d2
a = 8.99 x 109(2.5 x 10-12C)(4.2 x 10-3C)/((4.1 x 10-6 kg)(.12 m)2)
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Bri S.
asked 01/30/23A dust particle of mass 4.1 mg and charge +2.5 pC is released form rest 12 cm from a fixed charge of 4.2 mC. What will the dust particle's initial acceleration be when released?
a = F/m = keQ1Q2/d2
a = 8.99 x 109(2.5 x 10-12C)(4.2 x 10-3C)/((4.1 x 10-6 kg)(.12 m)2)
Please consider a tutor. Take care.
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