Question

Two point-charges Q1 and Q2 are 2.5 m apart, and their total charge is 19μC. If the force of repulsion between them is 0.07 N, what are magnitudes of the two charges by following the steps below?

a. Q1 + Q2 = _____________C
Use Coulomb's Law to calculate the product of the two charges.
b. Q1XQ2 = ________C
c. Solve for Q1 and Q2 using the two equations you got in part (a) and part (b). Hint: There are two possible solutions. Enter the answer with curly brackets as {Q1,Q2} where Q1 is the smaller charge and Q2 is the larger charge.
{Q1,Q2} = __________C

Answer 1

a)
`Q_1+Q_2=19* 10^(-6) C`

b)
`Q_1.Q_2=4.861* 10^(-11)\ C^2`

c)
`\{Q_1,Q_2\}=\{15.953* 10^(-6),3.047* 10^(-6)\}`

Step-by-step explanation:

Given:

• distance between the charges,
  `r=2.5\ m`

• total charge,
  `Q_1+Q_2=19* 10^(-6) C` .....................(1)

• repulsive force between the charges,
  `F=0.07\ N`

We first find the product of two charges using Coulomb's law:

`F=(1)/(4\pi.\epsilon_0)* (Q_1.Q_2)/(r^2)`

`0.07=9* 10^9* (Q_1.Q_2)/(2.5^2)`

`Q_1.Q_2=4.861* 10^(-11)\ C^2` ............................(2)

Now using eq.(1)&(2)

Put value of
`Q_1` from eq. (1) into eq. (2)

`(19* 10^(-6)-Q_2).Q_2=4.861* 10^(-11)\ C^2`

`(19* 10^(-6)-Q_2).Q_2=4.861* 10^(-11)`

`Q_2=15.953* 10^(-6)\ C\ or\ 3.047* 10^(-6)\ C`

Therefore, Charges:

`\{Q_1,Q_2\}=\{15.953* 10^(-6),3.047* 10^(-6)\}`