Question

A small metal bead, labeled A, has a charge of 26 nC . It is touched to metal bead B, initially neutral, so that the two beads share the 26 nC charge, but not necessarily equally. When the two beads are then placed 5.0 cm apart, the force between them is 4.4×10−4 N . Assume that A has a greater charge. Part A What is the charge qA and qB on the beads?

Answer 1

Answer:

$q_A=25.953\ \rm nC$

$q_B=0.047\ \rm nC$

Explanation:

Given:

• Total initial charge on bead A=26 nC
• The distance between them=5 cm
• Magnitude o the force between them $=4.4\times10^{-4}$

Using coulombs law the force between two charged particle is $\dfrac{q_Aq_B}{4\pi \epsilon_0 r^2}$

where r is the radial distance between them

According to question we have

$4.4\times10^{-4}=\dfrac{q_Aq_B\times9\times10^9}{0.05^2}\\\\4.4\times10^{-4}=\dfrac{q_A(q_A-26\times10^{-9})\times9\times10^9}{0.05^2}\\q_A=25.953\ \rm nC\\q_B=0.047\ \rm nC$

Hence the charge on two metal beads is calculated.