Question

Two charges of +8.0 mC and -6.0 mC attract each other with a force of 3.0 x 103 N in a vacuum. What is the distance between the charges?

Answer 1

Given :

• Two Charges Q1 = 8mC Q2 = -6mC
• Force between them = 3 × 10³ N

To Find :

• Distance between them (r)

Formula :

➞ Coulomb's Law -

$\bullet\underline{\boxed{\sf F = 9 \times 10^9\dfrac{Q_1Q_2}{r^2}}}$

Solution :

$\implies{\sf 3 \times 10^3 =\dfrac{9 \times 10^9 \times 8 \times 10^{-3} \times 6 \times 10^{-3}}{r^2}}$

$\implies{\sf r^2=\dfrac{9 \times 10^9 \times 8 \times 10^{-3} \times 2 \times 10^{-3}}{10^3}}$

$\implies{\sf r^2=\dfrac{144 \times 10^3}{10^3}}$

$\implies{\sf r^2 = 144 }$

$\implies{\bf r = 12\: m }$

Answer :

Distance between the charges is 12m

Answer 2

Given ,

➡ Two charges 8 mC and -6 mC attract each other with a force 3 × (10)³ N

We know that ,

The force between two charges is given by

$\mathtt{ \large\fbox{F =k \frac{ q_{1}q_{2} }{ {(r)}^{2} } }}$

Thus ,

$\sf \hookrightarrow 3 × {(10)}^{3} = \frac{9 \times {(10)}^{9} \times 8 \times {(10)}^{ - 3} \times 6 \times {(10)}^{ - 3} }{ {(r)}^{2} } \\ \\\sf \hookrightarrow {(r)}^{2} = \frac{9 \times 8 \times \cancel{6}\times \cancel{{(10)}^{3} }}{ \cancel{3} \times \cancel{ {(10)}^{3} }} \\ \\ \sf \hookrightarrow {(r)}^{2} = 9 \times 8 \times 2 \\ \\\sf \hookrightarrow {(r)}^{2} = 144 \\ \\\sf \hookrightarrow r = ± \: 12 \: m$

Since , the distance can't be negative

Hence , the distance between two charges is 12 m