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Question

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A 2.00 g plastic bead charged to -4.00 nC and a 4.00 g glass bead charged to +8.00 nC are 2.00 cm apart (center to center). What are the initial accelerations of (a) the plastic bead and (b) the glass bead?

Explanation / Answer

Start with F = kqq/r^2

F = (9 X 109)(4 X 10-9)(8 X 10-9)/.022

F = 7.2 X 10-4 N

F = ma

Part A) The plastic bead

7.2 X 10.-4 = .002(a)

a = .36 m/s2

Part B) The glass bead

7.2 X 10-4 = .004(a)

a = .18 m/s2

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