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PSYCHO15rus [73]
3 years ago
5

A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads sh

are the 28 nC charge, but not necessarily equally. When the two beads are then placed 5.0 cmapart, the force between them is 4.8×10^−4 N . Assume that A has a greater charge. What is the charge qA and qB on the beads?
Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
3 0

Answer:

Explanation:

Let the charge on bead A be q nC  and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = \frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

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