Question

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?

Answer 1

qA = 22.25 x 10^(-9) C

qB = 6.75 x 10^(9) C

Step-by-step 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

Answer 2

q(a) = 15 nC

q(b) = 10 nC

Step-by-step explanation:

The distance between q(a) and q(b) is 5 cm

Also, q(a) + q(b) = 29 nC

Force between them, F(a,b) = 5.4*10^-4 N

Using Coulomb's law

F = kq1q2 / r², where

F = force between the charges

K = constant of proportionality

q1 = magnitude of first charge

q2 = magnitude of the second charge

r = distance between the charges

5.4*10^-4 = {8.99*10^9 * q(a)[25 - q(a)] *10^-18} / 0.05²

5.4*10^-4 = {8.99*10^9 * q(a)[25 - q(a)] *10^-18} / 0.0025

{8.99*10^9 * q(a)[25 - q(a)] *10^-18} = 5.4*10^-4 * 0.0025

{8.99*10^9 * q(a)[25 - q(a)] *10^-18} = 1.35*10^-6

q(a)[25 - q(a)] *10^-18 = 1.36*10^-6 / 8.99*10^9

q(a)[25 - q(a)] *10^-18 = 1.50*10^-16

q(a)[25 - q(a)] = 1.50*10^-16 * 10^18

q(a)[25 - q(a)] = 150

25q(a) - q(a)² = 150

q(a)² - 25q(a) + 150 = 0 solve using quadratic equation, we have

q(a) - 10 = 0

q(a) - 15 = 0

Since the question stated that q(a) had the higher charge, then q(a) = 15 nC and q(b) = 10 nC