A small metal bead, labeled A, has a charge of 29 nC . It is touched to metal bead B, initially neutral, so that the two beads share the 29 nC charge, but not necessarily equally. When the two beads are then placed 5.0 cm apart, the force between them is 5.4×10−4 N . Assume that A has a greater charge. What are the charges qa and qb on the beads?

qA = 22.25 x 10^(-9) C

qB = 6.75 x 10^(9) C

Explanation:

We are given;

Sum of two charges = 29 nC

Thus, qA + qB = 29 x 10^(-9) C and

qA = 29 x 10^(-9)] - q2

Distance between 2 charges; r = 5cm = 0.05 m

Force between the 2 charges, F_(a,b) = 5.4 x 10^(-4) N

Now, from Coulomb's law, Force between charges is given as;

F = k•q1•q2/r², where

F = force between the charges

K = coulombs constant and has a value of 8.99 x 10^(9) N.m²/C²

q1 = qA= magnitude of charge A

q2 = qB = magnitude of charge B

r = distance between the charges

From earlier, we saw that;

qA = [29 x 10^(-9)] - qB

Plugging in the relevant values into the force equation, we have;

5.4 x 10^(-4) = [8.99 x 10^(9)•qA•qB]/0.05²

From earlier, we saw that;

qA = 29 x 10^(-9)] - qB

Thus,

5.4 x 10^(-4) = [8.99 x 10^(9)•(29 x 10^(-9)] - qB)•qB]/0.05²

5.4 x 10^(-4) = [8.99 x 10^(9)•(29 x 10^(-9)q2] - (qB)²)]/0.05²

5.4 x 10^(-4) x 0.05² = 260.71qB - 8.99 x 10^(9)(qB)²

8.99 x 10^(9)(qB)² - 260.71qB + 0.00000135 = 0

Finding the roots, we have;

qB = 6.75 x 10^(9) C or 22.25 x 10^(9) C

We know that;

qA = 29 x 10^(-9)] - qB

And from the question, we are told that qA has the greater charge, thus, we will use qB = 6.75 x 10^(9) C

So,

qA =

[29 x 10^(-9)] - 6.75 x 10^(9) = 22.25 x 10^(-9) C

explanation:

newton's third law, for every action there is an equal and opposite reaction.