Question

Two charges 2.0 × 10⁻⁶ C and 1.0 × 10⁻⁶ C are placed at a separation of 10 cm. Where should a third charge be placed, such that it experiences no net force due to these charges?

Answer 1

Given:

\[q_1  = 2 . 0 \times  {10}^{- 6}   C\]  

\[ q_2  = 1 . 0 \times  {10}^{- 6}   C\]

Let the third charge, q, be placed at a distance of x cm from charge q_1, as shown in the figure.  

 

 

By Coulomb's Law, force,      

     

\[F = \frac{1}{4\pi \epsilon_0}\frac{Q_1 Q_2}{r^2}\]

Force on charge q due to q₁,

\[F = \frac{9 \times {10}^9 \times 2 . 0 \times {10}^{- 6} \times q}{x^2}\]

Force on charge q due to q₂,

\[F' = \frac{9 \times {10}^9 \times {10}^{- 6} \times q}{\left( 10 - x \right)^2}\]

According to the question,

\[F - F' = 0\]

\[\Rightarrow F = F'\]
\[ \Rightarrow \frac{9 \times {10}^9 \times 2 \times {10}^{- 6} \times q}{x^2} = \frac{9 \times {10}^9 \times {10}^{- 6} \times q}{\left( 10 - x \right)^2}\]
\[ \Rightarrow x^2 = 2 \left( 10 - x \right)^2 \]
\[ \Rightarrow x^2 - 40x + 200 = 0\]
\[ \Rightarrow x = 20 \pm 10\sqrt{2}\]
\[ \Rightarrow x = 5 . 9 \text{ cm } ( \because x \neq 20 + 10\sqrt{2})\]

So, the third charge should be placed at a distance of 5.9 cm from q₁.