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

Answer:

a) $Q_1+Q_2=19\times 10^{-6} C$

b) $Q_1.Q_2=4.861\times 10^{-11}\ C^2$

c) $\{Q_1,Q_2\}=\{15.953\times 10^{-6},3.047\times 10^{-6}\}$

Explanation:

Given:

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

• total charge, $Q_1+Q_2=19\times 10^{-6} C$ .....................(1)

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

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

$F=\frac{1}{4\pi.\epsilon_0}\times \frac{Q_1.Q_2}{r^2}$

$0.07=9\times 10^9\times \frac{Q_1.Q_2}{2.5^2}$

$Q_1.Q_2=4.861\times 10^{-11}\ C^2$ ............................(2)

Now using eq.(1)&(2)

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

$(19\times 10^{-6}-Q_2).Q_2=4.861\times 10^{-11}\ C^2$

$(19\times 10^{-6}-Q_2).Q_2=4.861\times 10^{-11}$

$Q_2=15.953\times 10^{-6}\ C\ or\ 3.047\times 10^{-6}\ C$

Therefore, Charges:

$\{Q_1,Q_2\}=\{15.953\times 10^{-6},3.047\times 10^{-6}\}$